Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ If $a, b$ are positive real numbers and $a+b = 1$, prove that :
$$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$
I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.
Thank you.

Let $f(x)=\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2$. Note that $f(x)$ is convex on the $(0;+\infty)$ because its second derivative is equal to 
$$
f''(x)=2+\frac{6}{x^4}>0,~\text{for all}~x\in(0;+\infty).
$$
Hence, by Jensen's inequality ($p+q=1$)
$$
f(p)+f(q)\geqslant 2f\left(\frac{p+q}{2}\right)=2f\left(\frac{1}{2}\right)=\frac{25}{2}.
$$
Hence, 
$$
\left(p+\frac{1}{p}\right)^2+\left(q+\frac{1}{q}\right)^2\geq \frac{25}{2}.
$$
Thus, the correct answer is C.
Here is an elementary approach. We will prove that for $p\in(0;1)$ the following inequality holds
$$
\left(p+\frac{1}{p}\right)^2+\left(1-p+\frac{1}{1-p}\right)^2\geq \frac{25}{2},
$$
or
$$
p^2+(1-p)^2+\frac{1}{p^2}+\frac{1}{(1-p)^2}\geq \frac{17}{2},
$$
or
$$
(p^2+(1-p)^2)\left(\frac{p^2(1-p)^2+1}{p^2(1-p)^2}\right)\geq \frac{17}{2}.
$$
Now, note that 
$$
p^2+(1-p)^2=(p+(1-p))^2-2p(1-p)=1-2p(1-p).
$$ 
Denote $s=p(1-p)$. Thus, we need to prove that if $p\in(0;1)$ then
$$
(1-2s)\frac{s^2+1}{s^2}\geq\frac{17}{2},
$$
or
$$
2(1-2s)(s^2+1)\geq 17s^2.
$$
The last inequality is equivalent to 
$$
4s^3+15s^2+4s-1\leq 0.
$$
However, $4s^3+15s^2+4s-2=(4s-1)(s^2+4s+2)\leq 0$ because $s=p(1-p)\leq\frac{1}{4}$. 
