Mixing
Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$
$begingroup$
Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$
My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$
And from that we get:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
inequality proof-writing
edited Jan 3 at 23:59
Theo Bendit
17.2k12149
asked Jan 3 at 23:50
John D.John D.
153
$endgroup$
add a comment |
$begingroup$
Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$
My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$
And from that we get:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
inequality proof-writing
edited Jan 3 at 23:59
Theo Bendit
17.2k12149
asked Jan 3 at 23:50
John D.John D.
153
$endgroup$
$begingroup$
Use AM-GM inequality, like here
$endgroup$
– rtybase
Jan 3 at 23:53
$begingroup$
I don't think thea.m.-g.m.-inequalitytag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
$endgroup$
– Theo Bendit
Jan 3 at 23:59
$begingroup$
I know the AM-GM inequality but am not allowed to use it
$endgroup$
– John D.
Jan 4 at 0:00
$begingroup$
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
$endgroup$
– rtybase
Jan 4 at 0:05
add a comment |
$begingroup$
Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$
My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$
And from that we get:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
inequality proof-writing
edited Jan 3 at 23:59
Theo Bendit
17.2k12149
asked Jan 3 at 23:50
John D.John D.
153
$endgroup$
Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$
My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$
And from that we get:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
inequality proof-writing
inequality proof-writing
edited Jan 3 at 23:59
Theo Bendit
17.2k12149
asked Jan 3 at 23:50
John D.John D.
153
edited Jan 3 at 23:59
Theo Bendit
17.2k12149
asked Jan 3 at 23:50
John D.John D.
153
edited Jan 3 at 23:59
Theo Bendit
17.2k12149
edited Jan 3 at 23:59
Theo Bendit
17.2k12149
edited Jan 3 at 23:59
Theo Bendit
17.2k12149
17.2k12149
asked Jan 3 at 23:50
John D.John D.
153
asked Jan 3 at 23:50
John D.John D.
153
asked Jan 3 at 23:50
John D.John D.
153
153
$begingroup$
Use AM-GM inequality, like here
$endgroup$
– rtybase
Jan 3 at 23:53
$begingroup$
I don't think thea.m.-g.m.-inequalitytag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
$endgroup$
– Theo Bendit
Jan 3 at 23:59
$begingroup$
I know the AM-GM inequality but am not allowed to use it
$endgroup$
– John D.
Jan 4 at 0:00
$begingroup$
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
$endgroup$
– rtybase
Jan 4 at 0:05
add a comment |
$begingroup$
Use AM-GM inequality, like here
$endgroup$
– rtybase
Jan 3 at 23:53
$begingroup$
I don't think thea.m.-g.m.-inequalitytag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
$endgroup$
– Theo Bendit
Jan 3 at 23:59
$begingroup$
I know the AM-GM inequality but am not allowed to use it
$endgroup$
– John D.
Jan 4 at 0:00
$begingroup$
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
$endgroup$
– rtybase
Jan 4 at 0:05
$begingroup$
Use AM-GM inequality, like here
$endgroup$
– rtybase
Jan 3 at 23:53
$begingroup$
Use AM-GM inequality, like here
$endgroup$
– rtybase
Jan 3 at 23:53
$begingroup$
I don't think the
a.m.-g.m.-inequality tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.$endgroup$
– Theo Bendit
Jan 3 at 23:59
$begingroup$
I don't think the
a.m.-g.m.-inequality tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.$endgroup$
– Theo Bendit
Jan 3 at 23:59
$begingroup$
I know the AM-GM inequality but am not allowed to use it
$endgroup$
– John D.
Jan 4 at 0:00
$begingroup$
I know the AM-GM inequality but am not allowed to use it
$endgroup$
– John D.
Jan 4 at 0:00
$begingroup$
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
$endgroup$
– rtybase
Jan 4 at 0:05
$begingroup$
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
$endgroup$
– rtybase
Jan 4 at 0:05
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
From here:
$$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
move the $a$ to the left side to yield
$$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
The left side is a perfect square:
$$left(frac{x}{2} - frac{a}{2x}right)^2.$$
answered Jan 3 at 23:53
Theo BenditTheo Bendit
17.2k12149
$endgroup$
$begingroup$
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
$endgroup$
– Clayton
Jan 4 at 0:40
add a comment |
$begingroup$
Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$
answered Jan 3 at 23:54
Michael RozenbergMichael Rozenberg
98.5k1590189
$endgroup$
add a comment |
$begingroup$
$(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$
answered Jan 3 at 23:55
John_WickJohn_Wick
1,486111
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From here:
$$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
move the $a$ to the left side to yield
$$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
The left side is a perfect square:
$$left(frac{x}{2} - frac{a}{2x}right)^2.$$
answered Jan 3 at 23:53
Theo BenditTheo Bendit
17.2k12149
$endgroup$
$begingroup$
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
$endgroup$
– Clayton
Jan 4 at 0:40
add a comment |
$begingroup$
From here:
$$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
move the $a$ to the left side to yield
$$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
The left side is a perfect square:
$$left(frac{x}{2} - frac{a}{2x}right)^2.$$
answered Jan 3 at 23:53
Theo BenditTheo Bendit
17.2k12149
$endgroup$
$begingroup$
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
$endgroup$
– Clayton
Jan 4 at 0:40
add a comment |
$begingroup$
From here:
$$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
move the $a$ to the left side to yield
$$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
The left side is a perfect square:
$$left(frac{x}{2} - frac{a}{2x}right)^2.$$
answered Jan 3 at 23:53
Theo BenditTheo Bendit
17.2k12149
$endgroup$
From here:
$$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
move the $a$ to the left side to yield
$$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
The left side is a perfect square:
$$left(frac{x}{2} - frac{a}{2x}right)^2.$$
answered Jan 3 at 23:53
Theo BenditTheo Bendit
17.2k12149
answered Jan 3 at 23:53
Theo BenditTheo Bendit
17.2k12149
answered Jan 3 at 23:53
Theo BenditTheo Bendit
17.2k12149
answered Jan 3 at 23:53
Theo BenditTheo Bendit
17.2k12149
17.2k12149
$begingroup$
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
$endgroup$
– Clayton
Jan 4 at 0:40
add a comment |
$begingroup$
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
$endgroup$
– Clayton
Jan 4 at 0:40
$begingroup$
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
$endgroup$
– Clayton
Jan 4 at 0:40
$begingroup$
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
$endgroup$
– Clayton
Jan 4 at 0:40
add a comment |
$begingroup$
Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$
answered Jan 3 at 23:54
Michael RozenbergMichael Rozenberg
98.5k1590189
$endgroup$
add a comment |
$begingroup$
Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$
answered Jan 3 at 23:54
Michael RozenbergMichael Rozenberg
98.5k1590189
$endgroup$
add a comment |
$begingroup$
Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$
answered Jan 3 at 23:54
Michael RozenbergMichael Rozenberg
98.5k1590189
$endgroup$
Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$
answered Jan 3 at 23:54
Michael RozenbergMichael Rozenberg
98.5k1590189
answered Jan 3 at 23:54
Michael RozenbergMichael Rozenberg
98.5k1590189
answered Jan 3 at 23:54
Michael RozenbergMichael Rozenberg
98.5k1590189
answered Jan 3 at 23:54
Michael RozenbergMichael Rozenberg
98.5k1590189
98.5k1590189
add a comment |
add a comment |
$begingroup$
$(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$
answered Jan 3 at 23:55
John_WickJohn_Wick
1,486111
$endgroup$
add a comment |
$begingroup$
$(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$
answered Jan 3 at 23:55
John_WickJohn_Wick
1,486111
$endgroup$
add a comment |
$begingroup$
$(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$
answered Jan 3 at 23:55
John_WickJohn_Wick
1,486111
$endgroup$
$(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$
answered Jan 3 at 23:55
John_WickJohn_Wick
1,486111
answered Jan 3 at 23:55
John_WickJohn_Wick
1,486111
answered Jan 3 at 23:55
John_WickJohn_Wick
1,486111
answered Jan 3 at 23:55
John_WickJohn_Wick
1,486111
1,486111
add a comment |
add a comment |
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$begingroup$
Use AM-GM inequality, like here
$endgroup$
– rtybase
Jan 3 at 23:53
$begingroup$
I don't think the
a.m.-g.m.-inequalitytag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.$endgroup$
– Theo Bendit
Jan 3 at 23:59
$begingroup$
I know the AM-GM inequality but am not allowed to use it
$endgroup$
– John D.
Jan 4 at 0:00
$begingroup$
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
$endgroup$
– rtybase
Jan 4 at 0:05