Mixing
Proving $int_a^bsqrt{f^2+g^2}gesqrt{(int_a^b f)^2 + (int_a^b g)^2 } $ without using $|int_a^b...
$begingroup$
I have tried to prove this inequality
$$ int_a^b sqrt{f^2+g^2} ge sqrt{ left( int_a^b f right)^2 + left( int_a^b g right)^2 } $$
without the use the following theorem in complex analysis
$$left| int_a^b z(t),dt right| le int_a^b |z(t)|,dt$$
I failed. Can You help me with it?
real-analysis calculus complex-analysis analysis inequality
edited Jan 13 at 0:10
Blue
48.4k870154
asked Jan 12 at 21:35
Dr.AKADr.AKA
16610
$endgroup$
add a comment |
$begingroup$
I have tried to prove this inequality
$$ int_a^b sqrt{f^2+g^2} ge sqrt{ left( int_a^b f right)^2 + left( int_a^b g right)^2 } $$
without the use the following theorem in complex analysis
$$left| int_a^b z(t),dt right| le int_a^b |z(t)|,dt$$
I failed. Can You help me with it?
real-analysis calculus complex-analysis analysis inequality
edited Jan 13 at 0:10
Blue
48.4k870154
asked Jan 12 at 21:35
Dr.AKADr.AKA
16610
$endgroup$
$begingroup$
Please be more specific and explain what all your variables are and where they come from.
$endgroup$
– Viktor Glombik
Jan 12 at 21:45
$begingroup$
What tools do you have at your disposal? Are you familiar with the fact that for a real valued function $f$, we have $$left| int f right| le int |f|? $$
$endgroup$
– Xander Henderson
Jan 12 at 21:46
add a comment |
$begingroup$
I have tried to prove this inequality
$$ int_a^b sqrt{f^2+g^2} ge sqrt{ left( int_a^b f right)^2 + left( int_a^b g right)^2 } $$
without the use the following theorem in complex analysis
$$left| int_a^b z(t),dt right| le int_a^b |z(t)|,dt$$
I failed. Can You help me with it?
real-analysis calculus complex-analysis analysis inequality
edited Jan 13 at 0:10
Blue
48.4k870154
asked Jan 12 at 21:35
Dr.AKADr.AKA
16610
$endgroup$
I have tried to prove this inequality
$$ int_a^b sqrt{f^2+g^2} ge sqrt{ left( int_a^b f right)^2 + left( int_a^b g right)^2 } $$
without the use the following theorem in complex analysis
$$left| int_a^b z(t),dt right| le int_a^b |z(t)|,dt$$
I failed. Can You help me with it?
real-analysis calculus complex-analysis analysis inequality
real-analysis calculus complex-analysis analysis inequality
edited Jan 13 at 0:10
Blue
48.4k870154
asked Jan 12 at 21:35
Dr.AKADr.AKA
16610
edited Jan 13 at 0:10
Blue
48.4k870154
asked Jan 12 at 21:35
Dr.AKADr.AKA
16610
edited Jan 13 at 0:10
Blue
48.4k870154
edited Jan 13 at 0:10
Blue
48.4k870154
edited Jan 13 at 0:10
Blue
48.4k870154
48.4k870154
asked Jan 12 at 21:35
Dr.AKADr.AKA
16610
asked Jan 12 at 21:35
Dr.AKADr.AKA
16610
asked Jan 12 at 21:35
Dr.AKADr.AKA
16610
16610
$begingroup$
Please be more specific and explain what all your variables are and where they come from.
$endgroup$
– Viktor Glombik
Jan 12 at 21:45
$begingroup$
What tools do you have at your disposal? Are you familiar with the fact that for a real valued function $f$, we have $$left| int f right| le int |f|? $$
$endgroup$
– Xander Henderson
Jan 12 at 21:46
add a comment |
$begingroup$
Please be more specific and explain what all your variables are and where they come from.
$endgroup$
– Viktor Glombik
Jan 12 at 21:45
$begingroup$
What tools do you have at your disposal? Are you familiar with the fact that for a real valued function $f$, we have $$left| int f right| le int |f|? $$
$endgroup$
– Xander Henderson
Jan 12 at 21:46
$begingroup$
Please be more specific and explain what all your variables are and where they come from.
$endgroup$
– Viktor Glombik
Jan 12 at 21:45
$begingroup$
Please be more specific and explain what all your variables are and where they come from.
$endgroup$
– Viktor Glombik
Jan 12 at 21:45
$begingroup$
What tools do you have at your disposal? Are you familiar with the fact that for a real valued function $f$, we have $$left| int f right| le int |f|? $$
$endgroup$
– Xander Henderson
Jan 12 at 21:46
$begingroup$
What tools do you have at your disposal? Are you familiar with the fact that for a real valued function $f$, we have $$left| int f right| le int |f|? $$
$endgroup$
– Xander Henderson
Jan 12 at 21:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is quite easy. Let $c,d in mathbb R$ with $c^{2}+d^{2}=1$. Then $cint_a^{b}f +dint_a^{b} g=int_a^{b} (cf+dg)leq int_a^{b} sqrt {f^{2}+g^{2}}$ by C-S inequality. Now just take $c=frac {int_a^{b}f} {sqrt {(int_a^{b}f)^{2}+(int_a^{b}g)^{2}}}$ and $d=frac {int_a^{b}g} {sqrt {(int_a^{b}f)^{2}+(int_a^{b}g)^{2}}}$
answered Jan 13 at 0:05
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
where goes the $ b-a $ in your C-S inequality? : ) isn't that right that $ int_a^b 1 = b-a $ ?
$endgroup$
– Dr.AKA
Jan 15 at 4:21
$begingroup$
@Dr.AKA I am not applying C-S inequality in integral form. The only inequality needed here $|ax+by| leq sqrt {a^{2}+b^{2}} sqrt {x^{2}+y^{2}}$ for real numbers $a,b,x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:19
add a comment |
$begingroup$
Suppose $z(r)=f(r)+ig(r)$, where $r$ is a real variable and $f$ and $g$ are real valued functions. Then:
$(f(s)g(t)-f(t)g(s))^2geq0 quad;forall;;s,tinmathbb R implies f(s)^2g(t)^2+f(t)^2g(s)^2geq2f(t)g(t)f(s)g(s)\implies f(s)^2g(t)^2+f(s)^2f(t)^2+f(t)^2g(s)^2+g(t)^2g(s)^2geq f(s)^2f(t)^2+2f(t)g(t)f(s)g(s)+g(t)^2g(s)^2\implies(f(s)^2+g(s)^2)(f(t)^2+g(t)^2)geq(f(s)f(t)+g(t)g(s))^2\impliessqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}geq f(s)f(t)+g(t)g(s)$
Now, we'll need to use a generalized form of inequalities between functions for two variable functions. That'll give us (I'm going to assume $b>a$ in the following):
$$int_{s=a}^{s=b}int_{t=a}^{t=b}sqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}quad dtdsgeq int_{s=a}^{s=b}int_{t=a}^{t=b} (f(s)f(t)+g(t)g(s))quad dtds\impliesint_{s=a}^{s=b}int_{t=a}^{t=b}sqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}quad dtdsgeq int_{s=a}^{s=b}int_{t=a}^{t=b} (f(s)f(t))quad dsdtquad + int_{s=a}^{s=b}int_{t=a}^{t=b} (g(t)g(s))quad dsdt$$
These integrals may be simplified into:
$$Bigl(int_{r=a}^{r=b}sqrt{(f(r)^2+g(r)^2)}quad drBigr)^2geq Bigl(int_{r=a}^{r=b} f(r)quad drBigr)^2 + Bigl(int_{r=a}^{r=b} g(r)quad drBigr)^2$$
see this
So, finally, we have:
$$int_{r=a}^{r=b}sqrt{(f(r)^2+g(r)^2)}quad drgeq sqrt{Bigl(int_{r=a}^{r=b} f(r)quad drBigr)^2 + Bigl(int_{r=a}^{r=b} g(r)quad drBigr)^2}$$
Or:
$$int_a^b|z(r)|quad drgeq Biggl|int_a^bz(r)quad dr Biggr|$$
answered Jan 12 at 23:03
Cardioid_Ass_22Cardioid_Ass_22
36814
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add a comment |
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2 Answers
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active
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votes
2 Answers
2
active
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active
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$begingroup$
This is quite easy. Let $c,d in mathbb R$ with $c^{2}+d^{2}=1$. Then $cint_a^{b}f +dint_a^{b} g=int_a^{b} (cf+dg)leq int_a^{b} sqrt {f^{2}+g^{2}}$ by C-S inequality. Now just take $c=frac {int_a^{b}f} {sqrt {(int_a^{b}f)^{2}+(int_a^{b}g)^{2}}}$ and $d=frac {int_a^{b}g} {sqrt {(int_a^{b}f)^{2}+(int_a^{b}g)^{2}}}$
answered Jan 13 at 0:05
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
where goes the $ b-a $ in your C-S inequality? : ) isn't that right that $ int_a^b 1 = b-a $ ?
$endgroup$
– Dr.AKA
Jan 15 at 4:21
$begingroup$
@Dr.AKA I am not applying C-S inequality in integral form. The only inequality needed here $|ax+by| leq sqrt {a^{2}+b^{2}} sqrt {x^{2}+y^{2}}$ for real numbers $a,b,x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:19
add a comment |
$begingroup$
This is quite easy. Let $c,d in mathbb R$ with $c^{2}+d^{2}=1$. Then $cint_a^{b}f +dint_a^{b} g=int_a^{b} (cf+dg)leq int_a^{b} sqrt {f^{2}+g^{2}}$ by C-S inequality. Now just take $c=frac {int_a^{b}f} {sqrt {(int_a^{b}f)^{2}+(int_a^{b}g)^{2}}}$ and $d=frac {int_a^{b}g} {sqrt {(int_a^{b}f)^{2}+(int_a^{b}g)^{2}}}$
answered Jan 13 at 0:05
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
where goes the $ b-a $ in your C-S inequality? : ) isn't that right that $ int_a^b 1 = b-a $ ?
$endgroup$
– Dr.AKA
Jan 15 at 4:21
$begingroup$
@Dr.AKA I am not applying C-S inequality in integral form. The only inequality needed here $|ax+by| leq sqrt {a^{2}+b^{2}} sqrt {x^{2}+y^{2}}$ for real numbers $a,b,x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:19
add a comment |
$begingroup$
This is quite easy. Let $c,d in mathbb R$ with $c^{2}+d^{2}=1$. Then $cint_a^{b}f +dint_a^{b} g=int_a^{b} (cf+dg)leq int_a^{b} sqrt {f^{2}+g^{2}}$ by C-S inequality. Now just take $c=frac {int_a^{b}f} {sqrt {(int_a^{b}f)^{2}+(int_a^{b}g)^{2}}}$ and $d=frac {int_a^{b}g} {sqrt {(int_a^{b}f)^{2}+(int_a^{b}g)^{2}}}$
answered Jan 13 at 0:05
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
This is quite easy. Let $c,d in mathbb R$ with $c^{2}+d^{2}=1$. Then $cint_a^{b}f +dint_a^{b} g=int_a^{b} (cf+dg)leq int_a^{b} sqrt {f^{2}+g^{2}}$ by C-S inequality. Now just take $c=frac {int_a^{b}f} {sqrt {(int_a^{b}f)^{2}+(int_a^{b}g)^{2}}}$ and $d=frac {int_a^{b}g} {sqrt {(int_a^{b}f)^{2}+(int_a^{b}g)^{2}}}$
answered Jan 13 at 0:05
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
edited Jan 13 at 12:17
answered Jan 13 at 0:05
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
answered Jan 13 at 0:05
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
answered Jan 13 at 0:05
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
59.5k42161
$begingroup$
where goes the $ b-a $ in your C-S inequality? : ) isn't that right that $ int_a^b 1 = b-a $ ?
$endgroup$
– Dr.AKA
Jan 15 at 4:21
$begingroup$
@Dr.AKA I am not applying C-S inequality in integral form. The only inequality needed here $|ax+by| leq sqrt {a^{2}+b^{2}} sqrt {x^{2}+y^{2}}$ for real numbers $a,b,x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:19
add a comment |
$begingroup$
where goes the $ b-a $ in your C-S inequality? : ) isn't that right that $ int_a^b 1 = b-a $ ?
$endgroup$
– Dr.AKA
Jan 15 at 4:21
$begingroup$
@Dr.AKA I am not applying C-S inequality in integral form. The only inequality needed here $|ax+by| leq sqrt {a^{2}+b^{2}} sqrt {x^{2}+y^{2}}$ for real numbers $a,b,x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:19
$begingroup$
where goes the $ b-a $ in your C-S inequality? : ) isn't that right that $ int_a^b 1 = b-a $ ?
$endgroup$
– Dr.AKA
Jan 15 at 4:21
$begingroup$
where goes the $ b-a $ in your C-S inequality? : ) isn't that right that $ int_a^b 1 = b-a $ ?
$endgroup$
– Dr.AKA
Jan 15 at 4:21
$begingroup$
@Dr.AKA I am not applying C-S inequality in integral form. The only inequality needed here $|ax+by| leq sqrt {a^{2}+b^{2}} sqrt {x^{2}+y^{2}}$ for real numbers $a,b,x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:19
$begingroup$
@Dr.AKA I am not applying C-S inequality in integral form. The only inequality needed here $|ax+by| leq sqrt {a^{2}+b^{2}} sqrt {x^{2}+y^{2}}$ for real numbers $a,b,x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 5:19
add a comment |
$begingroup$
Suppose $z(r)=f(r)+ig(r)$, where $r$ is a real variable and $f$ and $g$ are real valued functions. Then:
$(f(s)g(t)-f(t)g(s))^2geq0 quad;forall;;s,tinmathbb R implies f(s)^2g(t)^2+f(t)^2g(s)^2geq2f(t)g(t)f(s)g(s)\implies f(s)^2g(t)^2+f(s)^2f(t)^2+f(t)^2g(s)^2+g(t)^2g(s)^2geq f(s)^2f(t)^2+2f(t)g(t)f(s)g(s)+g(t)^2g(s)^2\implies(f(s)^2+g(s)^2)(f(t)^2+g(t)^2)geq(f(s)f(t)+g(t)g(s))^2\impliessqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}geq f(s)f(t)+g(t)g(s)$
Now, we'll need to use a generalized form of inequalities between functions for two variable functions. That'll give us (I'm going to assume $b>a$ in the following):
$$int_{s=a}^{s=b}int_{t=a}^{t=b}sqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}quad dtdsgeq int_{s=a}^{s=b}int_{t=a}^{t=b} (f(s)f(t)+g(t)g(s))quad dtds\impliesint_{s=a}^{s=b}int_{t=a}^{t=b}sqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}quad dtdsgeq int_{s=a}^{s=b}int_{t=a}^{t=b} (f(s)f(t))quad dsdtquad + int_{s=a}^{s=b}int_{t=a}^{t=b} (g(t)g(s))quad dsdt$$
These integrals may be simplified into:
$$Bigl(int_{r=a}^{r=b}sqrt{(f(r)^2+g(r)^2)}quad drBigr)^2geq Bigl(int_{r=a}^{r=b} f(r)quad drBigr)^2 + Bigl(int_{r=a}^{r=b} g(r)quad drBigr)^2$$
see this
So, finally, we have:
$$int_{r=a}^{r=b}sqrt{(f(r)^2+g(r)^2)}quad drgeq sqrt{Bigl(int_{r=a}^{r=b} f(r)quad drBigr)^2 + Bigl(int_{r=a}^{r=b} g(r)quad drBigr)^2}$$
Or:
$$int_a^b|z(r)|quad drgeq Biggl|int_a^bz(r)quad dr Biggr|$$
answered Jan 12 at 23:03
Cardioid_Ass_22Cardioid_Ass_22
36814
$endgroup$
add a comment |
$begingroup$
Suppose $z(r)=f(r)+ig(r)$, where $r$ is a real variable and $f$ and $g$ are real valued functions. Then:
$(f(s)g(t)-f(t)g(s))^2geq0 quad;forall;;s,tinmathbb R implies f(s)^2g(t)^2+f(t)^2g(s)^2geq2f(t)g(t)f(s)g(s)\implies f(s)^2g(t)^2+f(s)^2f(t)^2+f(t)^2g(s)^2+g(t)^2g(s)^2geq f(s)^2f(t)^2+2f(t)g(t)f(s)g(s)+g(t)^2g(s)^2\implies(f(s)^2+g(s)^2)(f(t)^2+g(t)^2)geq(f(s)f(t)+g(t)g(s))^2\impliessqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}geq f(s)f(t)+g(t)g(s)$
Now, we'll need to use a generalized form of inequalities between functions for two variable functions. That'll give us (I'm going to assume $b>a$ in the following):
$$int_{s=a}^{s=b}int_{t=a}^{t=b}sqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}quad dtdsgeq int_{s=a}^{s=b}int_{t=a}^{t=b} (f(s)f(t)+g(t)g(s))quad dtds\impliesint_{s=a}^{s=b}int_{t=a}^{t=b}sqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}quad dtdsgeq int_{s=a}^{s=b}int_{t=a}^{t=b} (f(s)f(t))quad dsdtquad + int_{s=a}^{s=b}int_{t=a}^{t=b} (g(t)g(s))quad dsdt$$
These integrals may be simplified into:
$$Bigl(int_{r=a}^{r=b}sqrt{(f(r)^2+g(r)^2)}quad drBigr)^2geq Bigl(int_{r=a}^{r=b} f(r)quad drBigr)^2 + Bigl(int_{r=a}^{r=b} g(r)quad drBigr)^2$$
see this
So, finally, we have:
$$int_{r=a}^{r=b}sqrt{(f(r)^2+g(r)^2)}quad drgeq sqrt{Bigl(int_{r=a}^{r=b} f(r)quad drBigr)^2 + Bigl(int_{r=a}^{r=b} g(r)quad drBigr)^2}$$
Or:
$$int_a^b|z(r)|quad drgeq Biggl|int_a^bz(r)quad dr Biggr|$$
answered Jan 12 at 23:03
Cardioid_Ass_22Cardioid_Ass_22
36814
$endgroup$
add a comment |
$begingroup$
Suppose $z(r)=f(r)+ig(r)$, where $r$ is a real variable and $f$ and $g$ are real valued functions. Then:
$(f(s)g(t)-f(t)g(s))^2geq0 quad;forall;;s,tinmathbb R implies f(s)^2g(t)^2+f(t)^2g(s)^2geq2f(t)g(t)f(s)g(s)\implies f(s)^2g(t)^2+f(s)^2f(t)^2+f(t)^2g(s)^2+g(t)^2g(s)^2geq f(s)^2f(t)^2+2f(t)g(t)f(s)g(s)+g(t)^2g(s)^2\implies(f(s)^2+g(s)^2)(f(t)^2+g(t)^2)geq(f(s)f(t)+g(t)g(s))^2\impliessqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}geq f(s)f(t)+g(t)g(s)$
Now, we'll need to use a generalized form of inequalities between functions for two variable functions. That'll give us (I'm going to assume $b>a$ in the following):
$$int_{s=a}^{s=b}int_{t=a}^{t=b}sqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}quad dtdsgeq int_{s=a}^{s=b}int_{t=a}^{t=b} (f(s)f(t)+g(t)g(s))quad dtds\impliesint_{s=a}^{s=b}int_{t=a}^{t=b}sqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}quad dtdsgeq int_{s=a}^{s=b}int_{t=a}^{t=b} (f(s)f(t))quad dsdtquad + int_{s=a}^{s=b}int_{t=a}^{t=b} (g(t)g(s))quad dsdt$$
These integrals may be simplified into:
$$Bigl(int_{r=a}^{r=b}sqrt{(f(r)^2+g(r)^2)}quad drBigr)^2geq Bigl(int_{r=a}^{r=b} f(r)quad drBigr)^2 + Bigl(int_{r=a}^{r=b} g(r)quad drBigr)^2$$
see this
So, finally, we have:
$$int_{r=a}^{r=b}sqrt{(f(r)^2+g(r)^2)}quad drgeq sqrt{Bigl(int_{r=a}^{r=b} f(r)quad drBigr)^2 + Bigl(int_{r=a}^{r=b} g(r)quad drBigr)^2}$$
Or:
$$int_a^b|z(r)|quad drgeq Biggl|int_a^bz(r)quad dr Biggr|$$
answered Jan 12 at 23:03
Cardioid_Ass_22Cardioid_Ass_22
36814
$endgroup$
Suppose $z(r)=f(r)+ig(r)$, where $r$ is a real variable and $f$ and $g$ are real valued functions. Then:
$(f(s)g(t)-f(t)g(s))^2geq0 quad;forall;;s,tinmathbb R implies f(s)^2g(t)^2+f(t)^2g(s)^2geq2f(t)g(t)f(s)g(s)\implies f(s)^2g(t)^2+f(s)^2f(t)^2+f(t)^2g(s)^2+g(t)^2g(s)^2geq f(s)^2f(t)^2+2f(t)g(t)f(s)g(s)+g(t)^2g(s)^2\implies(f(s)^2+g(s)^2)(f(t)^2+g(t)^2)geq(f(s)f(t)+g(t)g(s))^2\impliessqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}geq f(s)f(t)+g(t)g(s)$
Now, we'll need to use a generalized form of inequalities between functions for two variable functions. That'll give us (I'm going to assume $b>a$ in the following):
$$int_{s=a}^{s=b}int_{t=a}^{t=b}sqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}quad dtdsgeq int_{s=a}^{s=b}int_{t=a}^{t=b} (f(s)f(t)+g(t)g(s))quad dtds\impliesint_{s=a}^{s=b}int_{t=a}^{t=b}sqrt{(f(s)^2+g(s)^2)}sqrt{(f(t)^2+g(t)^2)}quad dtdsgeq int_{s=a}^{s=b}int_{t=a}^{t=b} (f(s)f(t))quad dsdtquad + int_{s=a}^{s=b}int_{t=a}^{t=b} (g(t)g(s))quad dsdt$$
These integrals may be simplified into:
$$Bigl(int_{r=a}^{r=b}sqrt{(f(r)^2+g(r)^2)}quad drBigr)^2geq Bigl(int_{r=a}^{r=b} f(r)quad drBigr)^2 + Bigl(int_{r=a}^{r=b} g(r)quad drBigr)^2$$
see this
So, finally, we have:
$$int_{r=a}^{r=b}sqrt{(f(r)^2+g(r)^2)}quad drgeq sqrt{Bigl(int_{r=a}^{r=b} f(r)quad drBigr)^2 + Bigl(int_{r=a}^{r=b} g(r)quad drBigr)^2}$$
Or:
$$int_a^b|z(r)|quad drgeq Biggl|int_a^bz(r)quad dr Biggr|$$
answered Jan 12 at 23:03
Cardioid_Ass_22Cardioid_Ass_22
36814
edited Jan 13 at 9:46
answered Jan 12 at 23:03
Cardioid_Ass_22Cardioid_Ass_22
36814
answered Jan 12 at 23:03
Cardioid_Ass_22Cardioid_Ass_22
36814
answered Jan 12 at 23:03
Cardioid_Ass_22Cardioid_Ass_22
36814
36814
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$begingroup$
Please be more specific and explain what all your variables are and where they come from.
$endgroup$
– Viktor Glombik
Jan 12 at 21:45
$begingroup$
What tools do you have at your disposal? Are you familiar with the fact that for a real valued function $f$, we have $$left| int f right| le int |f|? $$
$endgroup$
– Xander Henderson
Jan 12 at 21:46