<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] If you know the Cauchy-Schwarz inequality, you can use it. Indeed, $LHS \ge \dfrac{1}{2}\left(p+\dfrac{1}{p}+q+\dfrac{1}{q}\right)^2= \dfrac{1}{2}\left(1+\dfrac{1^2}{p}+\dfrac{1^2}{q}\right)^2\ge\dfrac{1}{2}\left(1+\dfrac{(1+1)^2}{p+q}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2= \dfrac{25}{2} = RHS. $
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