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

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 $\text{RHS}=1-1/2+4+4=8.5$.
