Mixing
Proving $frac{a}{a^2-a+1}+frac{b}{b^2-b+1}+frac{c}{c^2-c+1}+frac{d}{d^2-d+1}le frac{8}{3}$ given $a+b+c+d=2$
$begingroup$
Let $a,b,c,din mathbb{R}$ and $a+b+c+d=2$. Prove that $$frac{a}{a^2-a+1}+frac{b}{b^2-b+1}+frac{c}{c^2-c+1}+frac{d}{d^2-d+1}le frac{8}{3}.$$
We have $$frac{a}{a^2-a+1}le frac{4}{3}aLongleftrightarrow frac{-aleft(2a-1right)^2}{3left(a^2-a+1right)}le 0,$$ (Right)
So $$LHSle frac{4}{3}left(a+b+c+dright)=frac{8}{3}=RHS$$
When $a=b=c=d=frac{1}{2}$
But also $a=b=c=1;d=-1$ i don't know how to solve it.
inequality
asked Feb 27 '18 at 11:39
Word ShallowWord Shallow
1,0832622
$endgroup$
add a comment |
$begingroup$
Let $a,b,c,din mathbb{R}$ and $a+b+c+d=2$. Prove that $$frac{a}{a^2-a+1}+frac{b}{b^2-b+1}+frac{c}{c^2-c+1}+frac{d}{d^2-d+1}le frac{8}{3}.$$
We have $$frac{a}{a^2-a+1}le frac{4}{3}aLongleftrightarrow frac{-aleft(2a-1right)^2}{3left(a^2-a+1right)}le 0,$$ (Right)
So $$LHSle frac{4}{3}left(a+b+c+dright)=frac{8}{3}=RHS$$
When $a=b=c=d=frac{1}{2}$
But also $a=b=c=1;d=-1$ i don't know how to solve it.
inequality
asked Feb 27 '18 at 11:39
Word ShallowWord Shallow
1,0832622
$endgroup$
1
$begingroup$
What have you attempted thus far?
$endgroup$
– user477343
Feb 27 '18 at 11:45
$begingroup$
@Word Please mention what you have already tried.
$endgroup$
– Fitz Watson
Feb 27 '18 at 12:20
$begingroup$
$$ f(a,b,c,d)=frac{1}{frac{(a-1)^2}{a}+1}+frac{1}{frac{(b-1)^2}{b}+1}+frac{1}{frac{(c-1)^2}{c}+1}+frac{1}{frac{(d-1)^2}{d}+1} $$
$endgroup$
– Amin235
Feb 27 '18 at 12:34
1
$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem, its source, the motivation behind the inequality, and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Carl Mummert
Feb 27 '18 at 13:19
add a comment |
$begingroup$
Let $a,b,c,din mathbb{R}$ and $a+b+c+d=2$. Prove that $$frac{a}{a^2-a+1}+frac{b}{b^2-b+1}+frac{c}{c^2-c+1}+frac{d}{d^2-d+1}le frac{8}{3}.$$
We have $$frac{a}{a^2-a+1}le frac{4}{3}aLongleftrightarrow frac{-aleft(2a-1right)^2}{3left(a^2-a+1right)}le 0,$$ (Right)
So $$LHSle frac{4}{3}left(a+b+c+dright)=frac{8}{3}=RHS$$
When $a=b=c=d=frac{1}{2}$
But also $a=b=c=1;d=-1$ i don't know how to solve it.
inequality
asked Feb 27 '18 at 11:39
Word ShallowWord Shallow
1,0832622
$endgroup$
Let $a,b,c,din mathbb{R}$ and $a+b+c+d=2$. Prove that $$frac{a}{a^2-a+1}+frac{b}{b^2-b+1}+frac{c}{c^2-c+1}+frac{d}{d^2-d+1}le frac{8}{3}.$$
We have $$frac{a}{a^2-a+1}le frac{4}{3}aLongleftrightarrow frac{-aleft(2a-1right)^2}{3left(a^2-a+1right)}le 0,$$ (Right)
So $$LHSle frac{4}{3}left(a+b+c+dright)=frac{8}{3}=RHS$$
When $a=b=c=d=frac{1}{2}$
But also $a=b=c=1;d=-1$ i don't know how to solve it.
inequality
inequality
asked Feb 27 '18 at 11:39
Word ShallowWord Shallow
1,0832622
asked Feb 27 '18 at 11:39
Word ShallowWord Shallow
1,0832622
edited Jan 29 at 11:48
Word Shallow
asked Feb 27 '18 at 11:39
Word ShallowWord Shallow
1,0832622
asked Feb 27 '18 at 11:39
Word ShallowWord Shallow
1,0832622
asked Feb 27 '18 at 11:39
Word ShallowWord Shallow
1,0832622
1,0832622
1
$begingroup$
What have you attempted thus far?
$endgroup$
– user477343
Feb 27 '18 at 11:45
$begingroup$
@Word Please mention what you have already tried.
$endgroup$
– Fitz Watson
Feb 27 '18 at 12:20
$begingroup$
$$ f(a,b,c,d)=frac{1}{frac{(a-1)^2}{a}+1}+frac{1}{frac{(b-1)^2}{b}+1}+frac{1}{frac{(c-1)^2}{c}+1}+frac{1}{frac{(d-1)^2}{d}+1} $$
$endgroup$
– Amin235
Feb 27 '18 at 12:34
1
$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem, its source, the motivation behind the inequality, and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Carl Mummert
Feb 27 '18 at 13:19
add a comment |
1
$begingroup$
What have you attempted thus far?
$endgroup$
– user477343
Feb 27 '18 at 11:45
$begingroup$
@Word Please mention what you have already tried.
$endgroup$
– Fitz Watson
Feb 27 '18 at 12:20
$begingroup$
$$ f(a,b,c,d)=frac{1}{frac{(a-1)^2}{a}+1}+frac{1}{frac{(b-1)^2}{b}+1}+frac{1}{frac{(c-1)^2}{c}+1}+frac{1}{frac{(d-1)^2}{d}+1} $$
$endgroup$
– Amin235
Feb 27 '18 at 12:34
1
$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem, its source, the motivation behind the inequality, and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Carl Mummert
Feb 27 '18 at 13:19
1
1
$begingroup$
What have you attempted thus far?
$endgroup$
– user477343
Feb 27 '18 at 11:45
$begingroup$
What have you attempted thus far?
$endgroup$
– user477343
Feb 27 '18 at 11:45
$begingroup$
@Word Please mention what you have already tried.
$endgroup$
– Fitz Watson
Feb 27 '18 at 12:20
$begingroup$
@Word Please mention what you have already tried.
$endgroup$
– Fitz Watson
Feb 27 '18 at 12:20
$begingroup$
$$ f(a,b,c,d)=frac{1}{frac{(a-1)^2}{a}+1}+frac{1}{frac{(b-1)^2}{b}+1}+frac{1}{frac{(c-1)^2}{c}+1}+frac{1}{frac{(d-1)^2}{d}+1} $$
$endgroup$
– Amin235
Feb 27 '18 at 12:34
$begingroup$
$$ f(a,b,c,d)=frac{1}{frac{(a-1)^2}{a}+1}+frac{1}{frac{(b-1)^2}{b}+1}+frac{1}{frac{(c-1)^2}{c}+1}+frac{1}{frac{(d-1)^2}{d}+1} $$
$endgroup$
– Amin235
Feb 27 '18 at 12:34
1
1
$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem, its source, the motivation behind the inequality, and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Carl Mummert
Feb 27 '18 at 13:19
$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem, its source, the motivation behind the inequality, and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Carl Mummert
Feb 27 '18 at 13:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, we simplify a bit by transforming $x = a-frac12, y = b - frac12, z = c - frac12, w = d - frac12$, so that we need to prove equivalently with $x+y+z+w=0$, the inequality
$$sum frac{2x+1}{4x^2+3} leqslant frac43$$
Here $sum$ represents cyclic sums. Also as $displaystyle 1 - 2cdotfrac{2x+1}{4x^2+3} = frac{(2x-1)^2}{4x^2+3}$, we may equivalently show
$$sum frac{(2x-1)^2}{4x^2+3} geqslant frac43$$
Now we need an upper bound on the denominator; for this, note
$$4x^2+3=3x^2+(y+z+w)^2+3leqslant 3x^2+3(y^2+z^2+w^2)+3 = 3sum x^2+3$$
Thus we have a common denominator now, therefore,
$$sum frac{(2x-1)^2}{4x^2+3} geqslant frac{sum (2x-1)^2}{3sum x^2+3} = frac{4sum x^2-4sum x+4}{3(sum x^2+1)}=frac43$$
answered Feb 27 '18 at 17:42
MacavityMacavity
35.7k52554
$endgroup$
$begingroup$
$(y+z+w)^2le 3left(y^2+z^2+w^2right)Leftrightarrow y=z=w$ So we have $z=x=w,x=y=w,x=y=zRightarrow x=y=z=w=1$
$endgroup$
– Word Shallow
Mar 9 '18 at 12:07
$begingroup$
but in that case, the equaLity occur when a=b=c=1,d=-1 and i can't soIve it
$endgroup$
– Word Shallow
Mar 9 '18 at 12:09
$begingroup$
Note the inequality is proven, you are looking for equality cases. In terms of $x,y,z,w$ equality is when $x=y=z=w=0$ and when $x,y,z=frac12, w=-frac32$ or permutations. Note in the latter case, numerators become zero, so the bounds on denominators are irrelevant.
$endgroup$
– Macavity
Mar 9 '18 at 12:27
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
First, we simplify a bit by transforming $x = a-frac12, y = b - frac12, z = c - frac12, w = d - frac12$, so that we need to prove equivalently with $x+y+z+w=0$, the inequality
$$sum frac{2x+1}{4x^2+3} leqslant frac43$$
Here $sum$ represents cyclic sums. Also as $displaystyle 1 - 2cdotfrac{2x+1}{4x^2+3} = frac{(2x-1)^2}{4x^2+3}$, we may equivalently show
$$sum frac{(2x-1)^2}{4x^2+3} geqslant frac43$$
Now we need an upper bound on the denominator; for this, note
$$4x^2+3=3x^2+(y+z+w)^2+3leqslant 3x^2+3(y^2+z^2+w^2)+3 = 3sum x^2+3$$
Thus we have a common denominator now, therefore,
$$sum frac{(2x-1)^2}{4x^2+3} geqslant frac{sum (2x-1)^2}{3sum x^2+3} = frac{4sum x^2-4sum x+4}{3(sum x^2+1)}=frac43$$
answered Feb 27 '18 at 17:42
MacavityMacavity
35.7k52554
$endgroup$
$begingroup$
$(y+z+w)^2le 3left(y^2+z^2+w^2right)Leftrightarrow y=z=w$ So we have $z=x=w,x=y=w,x=y=zRightarrow x=y=z=w=1$
$endgroup$
– Word Shallow
Mar 9 '18 at 12:07
$begingroup$
but in that case, the equaLity occur when a=b=c=1,d=-1 and i can't soIve it
$endgroup$
– Word Shallow
Mar 9 '18 at 12:09
$begingroup$
Note the inequality is proven, you are looking for equality cases. In terms of $x,y,z,w$ equality is when $x=y=z=w=0$ and when $x,y,z=frac12, w=-frac32$ or permutations. Note in the latter case, numerators become zero, so the bounds on denominators are irrelevant.
$endgroup$
– Macavity
Mar 9 '18 at 12:27
add a comment |
$begingroup$
First, we simplify a bit by transforming $x = a-frac12, y = b - frac12, z = c - frac12, w = d - frac12$, so that we need to prove equivalently with $x+y+z+w=0$, the inequality
$$sum frac{2x+1}{4x^2+3} leqslant frac43$$
Here $sum$ represents cyclic sums. Also as $displaystyle 1 - 2cdotfrac{2x+1}{4x^2+3} = frac{(2x-1)^2}{4x^2+3}$, we may equivalently show
$$sum frac{(2x-1)^2}{4x^2+3} geqslant frac43$$
Now we need an upper bound on the denominator; for this, note
$$4x^2+3=3x^2+(y+z+w)^2+3leqslant 3x^2+3(y^2+z^2+w^2)+3 = 3sum x^2+3$$
Thus we have a common denominator now, therefore,
$$sum frac{(2x-1)^2}{4x^2+3} geqslant frac{sum (2x-1)^2}{3sum x^2+3} = frac{4sum x^2-4sum x+4}{3(sum x^2+1)}=frac43$$
answered Feb 27 '18 at 17:42
MacavityMacavity
35.7k52554
$endgroup$
$begingroup$
$(y+z+w)^2le 3left(y^2+z^2+w^2right)Leftrightarrow y=z=w$ So we have $z=x=w,x=y=w,x=y=zRightarrow x=y=z=w=1$
$endgroup$
– Word Shallow
Mar 9 '18 at 12:07
$begingroup$
but in that case, the equaLity occur when a=b=c=1,d=-1 and i can't soIve it
$endgroup$
– Word Shallow
Mar 9 '18 at 12:09
$begingroup$
Note the inequality is proven, you are looking for equality cases. In terms of $x,y,z,w$ equality is when $x=y=z=w=0$ and when $x,y,z=frac12, w=-frac32$ or permutations. Note in the latter case, numerators become zero, so the bounds on denominators are irrelevant.
$endgroup$
– Macavity
Mar 9 '18 at 12:27
add a comment |
$begingroup$
First, we simplify a bit by transforming $x = a-frac12, y = b - frac12, z = c - frac12, w = d - frac12$, so that we need to prove equivalently with $x+y+z+w=0$, the inequality
$$sum frac{2x+1}{4x^2+3} leqslant frac43$$
Here $sum$ represents cyclic sums. Also as $displaystyle 1 - 2cdotfrac{2x+1}{4x^2+3} = frac{(2x-1)^2}{4x^2+3}$, we may equivalently show
$$sum frac{(2x-1)^2}{4x^2+3} geqslant frac43$$
Now we need an upper bound on the denominator; for this, note
$$4x^2+3=3x^2+(y+z+w)^2+3leqslant 3x^2+3(y^2+z^2+w^2)+3 = 3sum x^2+3$$
Thus we have a common denominator now, therefore,
$$sum frac{(2x-1)^2}{4x^2+3} geqslant frac{sum (2x-1)^2}{3sum x^2+3} = frac{4sum x^2-4sum x+4}{3(sum x^2+1)}=frac43$$
answered Feb 27 '18 at 17:42
MacavityMacavity
35.7k52554
$endgroup$
First, we simplify a bit by transforming $x = a-frac12, y = b - frac12, z = c - frac12, w = d - frac12$, so that we need to prove equivalently with $x+y+z+w=0$, the inequality
$$sum frac{2x+1}{4x^2+3} leqslant frac43$$
Here $sum$ represents cyclic sums. Also as $displaystyle 1 - 2cdotfrac{2x+1}{4x^2+3} = frac{(2x-1)^2}{4x^2+3}$, we may equivalently show
$$sum frac{(2x-1)^2}{4x^2+3} geqslant frac43$$
Now we need an upper bound on the denominator; for this, note
$$4x^2+3=3x^2+(y+z+w)^2+3leqslant 3x^2+3(y^2+z^2+w^2)+3 = 3sum x^2+3$$
Thus we have a common denominator now, therefore,
$$sum frac{(2x-1)^2}{4x^2+3} geqslant frac{sum (2x-1)^2}{3sum x^2+3} = frac{4sum x^2-4sum x+4}{3(sum x^2+1)}=frac43$$
answered Feb 27 '18 at 17:42
MacavityMacavity
35.7k52554
answered Feb 27 '18 at 17:42
MacavityMacavity
35.7k52554
answered Feb 27 '18 at 17:42
MacavityMacavity
35.7k52554
answered Feb 27 '18 at 17:42
MacavityMacavity
35.7k52554
35.7k52554
$begingroup$
$(y+z+w)^2le 3left(y^2+z^2+w^2right)Leftrightarrow y=z=w$ So we have $z=x=w,x=y=w,x=y=zRightarrow x=y=z=w=1$
$endgroup$
– Word Shallow
Mar 9 '18 at 12:07
$begingroup$
but in that case, the equaLity occur when a=b=c=1,d=-1 and i can't soIve it
$endgroup$
– Word Shallow
Mar 9 '18 at 12:09
$begingroup$
Note the inequality is proven, you are looking for equality cases. In terms of $x,y,z,w$ equality is when $x=y=z=w=0$ and when $x,y,z=frac12, w=-frac32$ or permutations. Note in the latter case, numerators become zero, so the bounds on denominators are irrelevant.
$endgroup$
– Macavity
Mar 9 '18 at 12:27
add a comment |
$begingroup$
$(y+z+w)^2le 3left(y^2+z^2+w^2right)Leftrightarrow y=z=w$ So we have $z=x=w,x=y=w,x=y=zRightarrow x=y=z=w=1$
$endgroup$
– Word Shallow
Mar 9 '18 at 12:07
$begingroup$
but in that case, the equaLity occur when a=b=c=1,d=-1 and i can't soIve it
$endgroup$
– Word Shallow
Mar 9 '18 at 12:09
$begingroup$
Note the inequality is proven, you are looking for equality cases. In terms of $x,y,z,w$ equality is when $x=y=z=w=0$ and when $x,y,z=frac12, w=-frac32$ or permutations. Note in the latter case, numerators become zero, so the bounds on denominators are irrelevant.
$endgroup$
– Macavity
Mar 9 '18 at 12:27
$begingroup$
$(y+z+w)^2le 3left(y^2+z^2+w^2right)Leftrightarrow y=z=w$ So we have $z=x=w,x=y=w,x=y=zRightarrow x=y=z=w=1$
$endgroup$
– Word Shallow
Mar 9 '18 at 12:07
$begingroup$
$(y+z+w)^2le 3left(y^2+z^2+w^2right)Leftrightarrow y=z=w$ So we have $z=x=w,x=y=w,x=y=zRightarrow x=y=z=w=1$
$endgroup$
– Word Shallow
Mar 9 '18 at 12:07
$begingroup$
but in that case, the equaLity occur when a=b=c=1,d=-1 and i can't soIve it
$endgroup$
– Word Shallow
Mar 9 '18 at 12:09
$begingroup$
but in that case, the equaLity occur when a=b=c=1,d=-1 and i can't soIve it
$endgroup$
– Word Shallow
Mar 9 '18 at 12:09
$begingroup$
Note the inequality is proven, you are looking for equality cases. In terms of $x,y,z,w$ equality is when $x=y=z=w=0$ and when $x,y,z=frac12, w=-frac32$ or permutations. Note in the latter case, numerators become zero, so the bounds on denominators are irrelevant.
$endgroup$
– Macavity
Mar 9 '18 at 12:27
$begingroup$
Note the inequality is proven, you are looking for equality cases. In terms of $x,y,z,w$ equality is when $x=y=z=w=0$ and when $x,y,z=frac12, w=-frac32$ or permutations. Note in the latter case, numerators become zero, so the bounds on denominators are irrelevant.
$endgroup$
– Macavity
Mar 9 '18 at 12:27
add a comment |
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$begingroup$
What have you attempted thus far?
$endgroup$
– user477343
Feb 27 '18 at 11:45
$begingroup$
@Word Please mention what you have already tried.
$endgroup$
– Fitz Watson
Feb 27 '18 at 12:20
$begingroup$
$$ f(a,b,c,d)=frac{1}{frac{(a-1)^2}{a}+1}+frac{1}{frac{(b-1)^2}{b}+1}+frac{1}{frac{(c-1)^2}{c}+1}+frac{1}{frac{(d-1)^2}{d}+1} $$
$endgroup$
– Amin235
Feb 27 '18 at 12:34
1
$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem, its source, the motivation behind the inequality, and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
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– Carl Mummert
Feb 27 '18 at 13:19