Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$
First $$(a+1/a)^2 + (b+1/b)^2 \geq\frac{1}{2} (a+b+1/a+1/b)^2=\frac{1}{2}(1+1/(ab))^2.$$
Then note that $$ab\le(a+b)^2/4=1/4.$$
Take it into the first one, you may get your inequality.
First Method.
$a^{2}+\dfrac{1}{a^{2}}\geq -15a+\dfrac{47}{4}$ $~$ $\Longleftrightarrow$ $~$ $(2a-1)^{2}(a^{2}+16a+4)\geq 0$ : evident
$\therefore$ $\left(a+\dfrac{1}{a}\right)^{2}+\left(b+\dfrac{1}{b}\right)^{2}=4+a^{2}+\dfrac{1}{a^{2}}+b^{2}+\dfrac{1}{b^{2}}\geq 4-15a+\dfrac{47}{4}-15b+\dfrac{47}{4}=\dfrac{25}{2}$
Second Method.
$\left(a+\dfrac{1}{a}\right)^{2}+\left(b+\dfrac{1}{b}\right)^{2}=4+\underline{a^{2}+b^{2}}+\underline{\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}}\geq 4+\underline{\dfrac{1}{2}}+\underline{8}=\dfrac{25}{2}$
- $(1^{2}+1^{2})(a^{2}+b^{2})\geq (a+b)^{2}$ : Cauchy-Schwarz
- $(a+b)(a+b)\left(\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}\right)\geq (1+1)^{3}$ : Holder
Generalization:
If $\sum_{1\le r\le n}a_r=S$ where $a_i$s are positive real numbers
$$\sum_{1\le r\le n}\left(a_r+\frac1{a_r}\right)^2=\sum_{1\le r\le n}a_r^2+\sum_{1\le r\le n}a_r^{-2}+2n$$
We know $$\frac{\sum_{1\le r\le n}a_r^m}n> \text{ or } <\left(\frac{\sum_{1\le r\le n}a_r}n\right)^m$$ according as $m$ lies or does not lie in $(0,1)$
Putting $m=2,$ $$\frac{\sum_{1\le r\le n}a_r^2}n>\left(\frac{\sum_{1\le r\le n}a_r}n\right)^2=\left(\frac Sn\right)^2=\frac{S^2}{n^2}$$
Putting $m=-2,$ $$\frac{\sum_{1\le r\le n}a_r^{-2}}n>\left(\frac{\sum_{1\le r\le n}a_r}n\right)^{-2}=\left(\frac Sn\right)^{-2}=\frac{n^2}{S^2}$$
On simplification, $$\sum_{1\le r\le n}\left(a_r+\frac1{a_r}\right)^2\ge \frac{(n^2+S^2)^2}{S^2n}$$
Here $S=1,n=2$