<s>[INST] 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.
 [/INST] Hint: It is $$\left(p+\frac{1}{p}\right)^2+\left(q+\frac{1}{q}\right)^2\geq \frac{25}{2}$$
This is equivalent to
$$(p^2+q^2)(1+\frac{1}{p^2q^2})\geq \frac{17}{2}$$
With $$p^2+q^2=1-2pq$$ we get
$$1-2pq+\frac{1}{p^2q^2}-\frac{2}{pq}\geq \frac{17}{2}$$ This is equivalent to
$$(1-4pq)(p^2q^2+4pq+2)\geq 0$$
And by AM-GM we get
$$\frac{p+q}{2}\geq \sqrt{pq}$$ or $$1-4pq\geq 0$$
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