Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$ 
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$

My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\\=a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+4\\=(a+b)^2-2ab+\bigg(\dfrac{1}{a}+\dfrac{1}{b}\bigg)^2-\dfrac{2}{ab}+4\\=4-2ab-\dfrac{2}{ab}+1+\dfrac{1}{a^2b^2}\\=4-2\bigg(\dfrac{a^2b^2+1}{ab}\bigg)+\dfrac{a^2b^2+1}{a^2b^2}\\=4-\bigg(\dfrac{a^2b^2+1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)\\=4-\bigg(ab+\dfrac{1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)$ 
Seems to using Cauchy-Schwartz. Please help.

Ok, so we have that $a+b=1$.
From AM-GM, this immediately tells us that $ab \leq 0.25$. Furthermore, since you said that $\frac{1}{a} + \frac{1}{b} = \frac{1}{ab}$, this shows that $\frac{1}{a} + \frac{1}{b} \geq 4$.
Now, we are given two numbers $a + \frac 1a$ and $b+ \frac 1b$. Their sum is greater than or equal to $5$, since $a+b=1$. Let their sum be $c$.
Then, by AM-GM, $(a+\frac{1}{a})(b+\frac 1b) \leq \frac{c^2}{4}$, whose minimum value is $6.25$, since $c \geq 5$.
Hence, it follows that $(a+\frac 1a)^2 + (b + \frac 1b)^2 = c^2 - 2(a+\frac{1}{a})(b+\frac 1b) \geq 25 - 12.5 \geq 12.5$, since $c \geq 5$.
This proof required nothing more than just ordinary AM-GM. Sometimes it's nice to think simple.
