<s>[INST] If $a,b>0$ and $a+b=1\;,$ Then minumum value of $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ is 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
 [/INST] 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 $\min\sum_{i=1}^nf(x_i) = nf\left(\frac{1}{n}\right)$.
</s>