<s>[INST] 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.
 [/INST] Without loss of generality, we can assume $a\ge b$, so let's write $a=2+x$, $b=2-x$ with $0\le x$.  The inequality is clearly satisfied if $a\ge4$, so we need only worry about the range $0\le x\lt2$.  Plugging this into the OP's inequality and simplifying like crazy, we find we need only prove
$$f(x)=x^2+{(4+x^2)\over(4-x^2)^2}\ge{1\over4}\quad\text{for }0\le x\lt2$$
Now it does not require calculus to see that the function $f(x)$ is increasing on the interval $0\le x\lt2$:  The $x^2$ term is clearly increasing, and so is the quotient term, since the numerator $4+x^2$ is increasing and the denominator $(4-x^2)^2$ is decreasing.  Finally, since $f(0)={1\over4}$, we can conclude that the inequality holds.
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