Prove a well-known inequality using Cauchy-Schwarz or AM-GM

Question

For $a;b>0$ and $ab \geq 1$ we have a well-known inequality:

$\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2} \geq \dfrac{2}{1+ab}$

Which is equivalent to: $\dfrac{(a-b)^2(ab-1)}{(1+a^2)(1+b^2)(1+ab)} \geq 0$ (true)

But is there a solution for this inequality by Cauchy-Schwarz or AM-GM? Thank you.

Answer 1

Again not AM-GM or C-S (unless you count $a^2+b^2\geq2ab$) but I thought this was kinda sweet: \begin{align*} \frac{1}{1+a^2}+\frac{1}{1+b^2}=1-\frac{(ab)^2-1}{1+a^2+b^2+(ab)^2}&\geq1-\frac{(ab)^2-1}{1+2ab+(ab)^2}\\ &=1-\frac{ab-1}{ab+1}\\ &=\frac{2}{1+ab}. \end{align*}

Answer 2

Edit: Came up with another solution that actually uses the AM-GM inequality.

We start with the initial inequality, and multiply the first two factions by $\frac{1}{2}b^2$ and $\frac{1}{2}a^2$ on the top and bottom respectively: $$\frac{\frac{1}{2}b^2}{\frac{b^2+(ab)^2}{2}}+\frac{\frac{1}{2}a^2}{\frac{a^2+(ab)^2}{2}}\geq\frac{2}{1+ab}$$

This sets us up to use the AM-GM inequality on the denominators of both (keeping in mind since $a$ and $b$ are always greater than zero this works): $$\frac{1}{2a}+\frac{1}{2b}\geq\frac{2}{1+ab}$$

We can tidy that up - rinse and repeat with the AM-GM inequality by dividing by 2: $$\frac{a+b}{2}(1+ab)\geq 2ab$$ $$(1+ab)\geq 2\sqrt{ab}$$

And again, to finally arrive that: $$\sqrt{ab}\geq\sqrt{ab}$$

Which is always true for $a,b>0$ and $ab\geq 1$. (Not a very technical answer, there are gaps, should really apply AM-GM to the smaller inequality, but hopefully that gets you started)

---

Probably kinda convoluted, but I've done this more algebraically.

Let $a=u+v$ and $b=u-v$, where $u>v$ by the restrictions given.

Expanding the expression produced we get the following: $$\frac{1}{1+u^2+2uv+v^2}+\frac{1}{1+u^2-2uv+v^2}\geq\frac{2}{1+u^2-v^2}$$

And then simplifying using difference of two squares: $$\frac{2(1+u^2+v^2)}{(1+u^2+v^2)^2-4(uv)^2}\geq\frac{2}{1+u^2-v^2}$$

Cross multiplying and simplifying again gets: $$(1+u^2)^2-(1+u^2+v^2)^2\geq v^4-4(uv)^2$$

And we expand and simplify: $$u^2-v^2\geq1$$

By difference of two squares, we can substitute $a$ and $b$ back in and we're left with $ab\geq1$, which is one of the restrictions given.

Edit - so I realised you wanted it via one of the inequalities directly, but this solution doesn't give it that way. However, I did take inspiration from the AM-GM inequality - if you notice, $u=\frac{a+b}{2}$ if you solve for $u$ independently, and $ab$ winds up being $u^2-v^2$, which ends up being a fairly helpful expression. It works out to be slightly more intuitive if you do the substitution for $a$ and $b$ rather than applying AM-GM directly (using this solution anyway). Hope this helps.