If $a,b\in\mathbb R\setminus\{0\}$ and $a+b=4$, prove that $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2\ge12.5$. If $a,b\in\mathbb R\setminus\{0\}$ and $a+b=4$, prove that $$\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge12.5$$
I could expand everything: $$a^2+2+\frac{1}{a^2}+b^2+2+\frac{1}{b^2}\ge12.5$$
Subtract $4$ from both sides: $$a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}\ge8.5$$
We could use AM-GM here ($a^2,\frac{1}{a^2},b^2,\frac{1}{b^2}$ are all positive), but obviously it wouldn't do anything useful. And we still have to use the fact that $a+b=4$ somehow.
I've tried substituting $b$ with $4-a$, but after clearing the denominators and simplifying we don't quite see anything useful, just a random 6th degree polynomial.
The polynomial is actually: $$2a^6-24 a^5+103.5 a^4-188 a^3+122 a^2-8 a+16\ge 0$$
How could I solve this? We can't use calculus by the way. Thanks.

You can do this with the quadratic-arithmetic mean: (this is possible, because $a^2\geq 0$.)
$$
\sqrt{\frac{a^2+b^2}2}\geq\frac{a+b}2\\
a^2+b^2\geq \frac{(a+b)^2}2=8
$$
Now, you only have to proof $\frac 1{a^2}+\frac 1{b^2}\geq \frac 12$.
Just as before, we know that
$$
\frac 1{a^2}+\frac 1{b^2}\geq \frac{(\frac 1a+\frac 1b)^2}2
$$
We know that the minimum of $\frac 1a+\frac 1{4-a}$ is at $a=2$, with outcome $1$. (This can be done by differentiation, or multiply with $a(4-a)$ first.) Now we get
$$
\frac 1{a^2}+\frac 1{b^2}\geq \frac{(\frac 1a+\frac 1b)^2}2\geq \frac 12
$$
