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If $a,b>0$ and $a+b=1\;,$ Then minumum value of $\displaystyle \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2$ is

$\bf{My\; Try::}$ Let $a=\sin^2 \theta$ and $b=\cos^2 \theta\;,$ Then We have to minimize

$$\displaystyle f(\theta) = (\sin^2 \theta+\csc^2 \theta)^2+(\cos^2 \theta+\sec ^2 \theta) = \sin^4 \theta+\csc^4 \theta+\cos^4 \theta+\sec^4 \theta+4$$

So $$f(\theta) = 1-2\sin^2 \theta\cos^2 \theta+\frac{1}{\sin^2 \theta \cos^2 \theta}+4$$

So $$f(\theta) = 1-\frac{\sin^2 2\theta}{2}+\frac{4}{\sin^2 \theta}+4$$

Now Put $\sin^2 2\theta = t\;, t\in \left(0,1\right]$

So we get $$f(t)=5-\frac{t}{2}+\frac{4}{t}\;,$$ So we get $\displaystyle f'(t) = -\frac{1}{2}-\frac{4}{t^2}<0\;\forall t \in (0,1]$

So $$f(1)_{\min} = 5-\frac{1}{2}+4=9-\frac{1}{2}=\frac{17}{2}$$

But answer given as $$\frac{25}{2}$$

plz help me, Where I am wrong, Thanks

juantheron
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    A simpler solution: $f(x) = \left(x + \frac{1}{x}\right)^2$ satisfy $f''(x) = 2 + \frac{6}{x^4} > 0$ so $f$ is convex on $[0,1]$ and therefore $$f(a) + f(b) \geq 2f\left(\frac{a+b}{2}\right) = 2f\left(\frac{1}{2}\right) = \frac{25}{2}$$ – Winther May 31 '16 at 05:22
  • @Winther I think your post could be posted as an answer. (A very elegant answer, if I may say so.) – Martin Sleziak May 31 '16 at 07:00
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    Now I found that there is another question about the same inequality: Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ – Martin Sleziak May 31 '16 at 07:57
  • juantheron: If the main point of your question is to discuss this particular attempt to prove the inequality, you should probably add the ([tag:proof-verification]) tag. If you are interested in any proof, then it is duplicate of the older question. (And even if it is proof verification question, if you are already satisfied with the comments you have received on your proof, I think that it can be closed as a duplicate. So that if other users are looking for this inequality and find this question, they are directed to the older one.) – Martin Sleziak May 31 '16 at 08:14
  • Excellent solution by Winther. but can i know how to prove equality occurs when $a=b=0.5$ – Umesh shankar Mar 11 '19 at 09:46

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The following is not true $$\sin^4 \theta+\csc^4 \theta+\cos^4 \theta+\sec^4 \theta+4=1-2\sin^2 \theta\cos^2 \theta+\frac{1}{\sin^2 \theta \cos^2 \theta}+4.$$

For example, let $\theta=\pi/4$. Then $\text{LHS}=1/4+4+1/4+4+4=12.5$, while $\text{RHS}=1-1/2+4+4=8.5$.

Eclipse Sun
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Where I am wrong

In the following part :

$$f(\theta) = 1-2\sin^2 \theta\cos^2 \theta+\frac{1}{\sin^2 \theta \cos^2 \theta}+4$$

This is not correct.

$$\begin{align}f(\theta)&=\sin^4\theta+\cos^4\theta+\frac{1}{\sin^4\theta}+\frac{1}{\cos^4\theta}+4\\&=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta+\frac{(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta}{\sin^4\theta\cos^4\theta}+4\\&=1-2\sin^2\theta\cos^2\theta+\frac{1-2\sin^2\theta\cos^2\theta}{\sin^4\theta\cos^4\theta}+4\end{align}$$

As you did, you can put $\sin^22\theta=t$ to see that $$f(t)=\frac{16}{t^2}-\frac t2-\frac 8t+5$$ is decreasing for $0\lt t\le 1$, so the answer is $f(1)=\frac{25}{2}$ as desired.

mathlove
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Here is an alternative solution: take $f(x) = \left(x + \frac{1}{x}\right)^2$ then $f''(x) = 2 + \frac{6}{x^4} > 0$ so $f$ is a convex function and therefore satisfy $$f(a) + f(b) \geq 2f\left(\frac{a+b}{2}\right) = 2f\left(\frac{1}{2}\right) = \frac{25}{2}$$

This argument can also be used to solve other similar types of problems. For example we have the following generalization: if $f:[0,1]\to\mathbb{R}$ is a convex function and $x_i\in[0,1]$ with $x_1+\ldots+x_n = 1$ then $\min\sum_{i=1}^nf(x_i) = nf\left(\frac{1}{n}\right)$.

Winther
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  • Can we use Jensen's inequality to prove equality condition that is Minimum occurs when $a=b=0.5$ – Umesh shankar Mar 11 '19 at 09:52
  • @Umeshshankar The inequality I use here is Jensens. – Winther Mar 11 '19 at 10:00
  • Yes i am asking Jensen's inequality is an equality when the curve is a straight line. But here the curve is not a line? – Umesh shankar Mar 11 '19 at 10:27
  • @Umeshshankar Jensen's inequality, $tf(a) + (1-t)f(b) \geq f(ta + (1-t)b)$, is an equality for all $t\in[0,1]$ only when the curve is a straight line. It can be an equality for a single value without it being a straight line. – Winther Mar 11 '19 at 11:54
  • ok do you mean to say $f(x)$ is a constant function – Umesh shankar Mar 11 '19 at 13:05
  • What are you talking about? I don't understand where you got that from and what you are trying to understand – Winther Mar 11 '19 at 13:06
  • Because if $f(x)=c$ is a constant function we have $f(a)=c=f(b)=f(ta+(1-t)b)$ Then we get $tc+(1-t)c \ge c$ which is actually an equality.So $\left(a+\frac{1}{a}\right)^2=\left(b+\frac{1}{b}\right)^2=c=\frac{25}{4}$ from which we get $a=b=0.5$ Am i right? – Umesh shankar Mar 11 '19 at 13:11
  • The function in question is not the constant function so that argument makes little sense – Winther Mar 11 '19 at 13:15
  • It can be an equality for a single value without it being a straight line. Can you explain this please – Umesh shankar Mar 11 '19 at 13:59