<s>[INST] If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that $(a+1/a)^2+(b+1/b)^2\ge 25/2$ 
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that 
  $$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.$$  

My work:
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge \dfrac{25}{2}\implies a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}+4\ge \dfrac{25}{2}$$
Now, we have $a^2+\dfrac{1}{a^2}\ge 2$ and $b^2+\dfrac{1}{b^2}\ge 2$.
Here, I am stuck, I cannot use the information provided, $a+b=1$ to any use. Please help!
 [/INST] Hint: Substitute $a=\frac{1}{2}+x$ and $b=\frac{1}{2}-x$, $|x|<\frac{1}{2}$. Then you should only find the minimum of a one variable function $f(x)$.
I'll write the whole solution, maybe someone finds it helpful. After the substitution we get:
$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2=\frac{9}{2}+2x^2+\frac{\frac{1}{2}+2x^2}{(\frac{1}{4}-x^2)^2}=f(x)$.
Since $x^2\geqslant0$, we get that $f(x)\geqslant\frac{9}{2}+\frac{\frac{1}{2}}{\frac{1}{16}}=\frac{25}{2}$. Equality: for $x=0$ or $a=b=\frac{1}{2}$.
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