If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that $(a+1/a)^2+(b+1/b)^2\ge 25/2$ [duplicate]

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!

Solution 1:

By QM-AM, $\displaystyle a^2+b^2 \geq \frac{1}{2}(a+b)^2 = \frac{1}{2}$. QM-AM again gives $\displaystyle \frac{1}{a^2}+\frac{1}{b^2} \geq \frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)^2$. By AM-HM, $\displaystyle \frac{2}{\frac{1}{a}+\frac{1}{b}} \leq \frac{a+b}{2} = \frac{1}{2}$, whence $\displaystyle\frac{1}{a}+\frac{1}{b} \geq 4$, and thus $\displaystyle \frac{1}{a^2}+\frac{1}{b^2} \geq 8$. Thus we have $$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 = a^2 + b^2 + \frac{1}{a^2}+\frac{1}{b^2} + 4 \geq \frac{1}{2} + 8 + 4 = \frac{25}{2}.$$

Solution 2:

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 $a=b=\frac{1}{2}$.