I want to prove for all $a > 0 $, $b > 0$, $\frac{2ab}{a + b} + \sqrt{\frac{1}{2}\left(a^2 + b^2\right)} \geq \frac{a + b}{2} + \sqrt{ab}$

Question

How can I prove that for all $a > 0$ and $b > 0$, $$\frac{2ab}{a + b} + \sqrt{\frac{1}{2}\left(a^2 + b^2\right)} \geq \frac{a + b}{2} + \sqrt{ab}$$ I found this inequality in an mathematic olympiade and I could not not prove it.

Answer 1

This is false in general.

$\dfrac{2ab}{a + b} + \sqrt{\frac{1}{2}\left(a^2 + b^2\right)} \geq \dfrac{a + b}{2} + \sqrt{ab} $

If $b=ca$ this is $\dfrac{2ca^2}{a(1+c)} + \sqrt{\frac{1}{2}\left(a^2(1+c^2)\right)} \geq \dfrac{a(1+c)}{2} + \sqrt{ca^2} $

or $\dfrac{2c}{1+c} + \sqrt{\frac{1}{2}\left(1+c^2\right)} \geq \dfrac{1+c}{2} + \sqrt{c} $ or $\dfrac{2c}{1+c}-\dfrac{1+c}{2} \geq \sqrt{c}-\sqrt{\frac{1}{2}\left(1+c^2\right)} $

or $\dfrac{(1-c)^2}{2(1+c)} \leq \sqrt{\frac{1}{2}\left(1+c^2\right)} -\sqrt{c} $

For $c=2$ this is $\dfrac{1}{4} \leq \sqrt{\frac{1}{2}\left(5\right)} -\sqrt{2} \approx 0.1669 $

which is false.

I suspected this result when I went through the following series of equivalences which ended with something that could be false.

$\dfrac{2ab}{a + b} -\dfrac{a + b}{2} \geq \sqrt{ab}-\sqrt{\frac{1}{2}\left(a^2 + b^2\right)} $

$\begin{array}\\ \dfrac{(a-b)^2}{2(a+b)} &\leq \sqrt{\frac{1}{2}\left(a^2 + b^2\right)}-\sqrt{ab}\\ &= (\sqrt{\frac{1}{2}\left(a^2 + b^2\right)}-\sqrt{ab})\dfrac{\sqrt{\frac{1}{2}\left(a^2 + b^2\right)}+\sqrt{ab}}{\sqrt{\frac{1}{2}\left(a^2 + b^2\right)}+\sqrt{ab}}\\ &= \dfrac{\frac{1}{2}\left(a^2 + b^2\right)-ab}{\sqrt{\frac{1}{2}\left(a^2 + b^2\right)}+\sqrt{ab}}\\ &= \dfrac{\frac{1}{2}(a-b)^2}{\sqrt{\frac{1}{2}\left(a^2 + b^2\right)}+\sqrt{ab}}\\ a+b &\ge \sqrt{\frac{1}{2}\left(a^2 + b^2\right)}+\sqrt{ab}\\ (a+b)^2 &\ge \frac{1}{2}\left(a^2 + b^2\right)+ab+2\sqrt{\frac{1}{2}\left(a^2 + b^2\right)ab}\\ a^2+2ab+b^2 &\ge \frac{1}{2}\left(a^2 + b^2+2ab\right)+2\sqrt{\frac{1}{2}\left(a^2 + b^2\right)ab}\\ \frac12(a^2+2ab+b^2) &\ge 2\sqrt{\frac{1}{2}\left(a^2 + b^2\right)ab}\\ \frac14(a+b)^2 &\ge \sqrt{2\left(a^2 + b^2\right)ab}\\ \frac1{16}(a+b)^4 &\ge 2\left(a^2 + b^2\right)ab\\ (a+b)^4 &\ge 32\left(a^2 + b^2\right)ab\\ \end{array} $

$\begin{array}\\ (a+b)^4 - 32\left(a^2 + b^2\right)ab &=a^4-28a^3b+6a^2b^2-28ab^3+b^4\\ &=a^4-4a^3b+6a^2b^2-4ab^3+b^4 -24(a^3b+ab^3)\\ &=(a-b)^4 -24ab(a+b)\\ \end{array} $

$\begin{array}\\ (a-b)^4 &=a^4-4a^3b+6a^2b^2-4ab^3+b^4\\ \end{array} $

Answer 2

Rearranging terms and simplifying a bit: $$ \sqrt{2\left(a^2 + b^2\right)} -2\sqrt{ab}\geq \frac{(a-b)^2}{a+b} $$ Multiply both sides by $\sqrt{2\left(a^2 + b^2\right)} +2\sqrt{ab}$:$$ 2\left(a^2 + b^2\right) -4ab\ge \frac{(a-b)^2}{a+b}(\sqrt{2\left(a^2 + b^2\right)} +2\sqrt{ab})\\ 2(a+b)\ge \sqrt{2\left(a^2 + b^2\right)} +2\sqrt{ab} $$ Square both sides, simplify, square again: $$ 4a^2 +8ab + 4b^2\ge 2a^2 + 4ab + 2 b^2 + 4\sqrt{2\left(a^2 + b^2\right)} \sqrt{ab}\\ (a+b)^2 \ge 2\sqrt{2ab\left(a^2 + b^2\right)}\\ (a+b)^4 \ge 8a^3b + 8ab^3 \\ a^4 - 4 a^3b + 6 a^2 b^2 - 4 ab^3 + b^4 \ge 0\\ (a-b)^4 \ge 0 $$

Answer 3

Note that we may assume $b = 1$. (Indeed, divide both sides by $b$ and substitute $a \mapsto ab$.) So, it suffices to establish the following inequality:

$$ \frac{2a}{a+1} + \sqrt{\frac{a^2+1}{2}} \ \stackrel{?}{\geq}\ \frac{a+1}{2} + \sqrt{a} \tag{1} $$

Now let $y = \frac{2\sqrt{a}}{1+a}$, or equivalently, $\frac{1}{y} = \frac{1}{2}(\sqrt{a} + \frac{1}{\sqrt{a}})$. Also, note that $0 \leq y \leq 1$. Then

\begin{align*} \text{(1)} &\stackrel{\div\sqrt{a}}{\qquad\iff\qquad} y + \sqrt{2y^{-2} - 1} \geq \frac{1}{y} + 1 \\[0.5em] &\stackrel{\times y}{\qquad\iff\qquad} \sqrt{2-y^2} \geq 1 + y - y^2 \\[0.5em] &\stackrel{\text{square}}{\qquad\iff\qquad} 2-y^2 \geq (1 + y - y^2)^2 \\[0.5em] &\qquad\iff\qquad 1 - 2 y + 2 y^3 - y^4 \geq 0 \\[0.5em] &\qquad\iff\qquad (1-y)^3(1+y) \geq 0, \end{align*}

which is true because $0 \leq y \leq 1$.

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Addendum. If we define the $p$-norm of the list $\mathbf{x} = (a, b)$ by

$$ \| \mathbf{x} \|_p = \begin{cases} \bigl( \frac{a^p + b^p}{2} \bigr)^{1/p}, & p \neq 0 \\[0.5em] \sqrt{ab}, & p = 0 \end{cases} $$

then $p \mapsto \| \mathbf{x} \|_p$ is an increasing continuous function and the inequality is recast as

$$ \| \mathbf{x} \|_{-1} + \| \mathbf{x} \|_{2} \geq \| \mathbf{x} \|_{1} + \| \mathbf{x} \|_{0}. $$

However, this interpretations doesn't seem providing any useful insight on the problem. (The inflection points of $p \mapsto \| \mathbf{x} \|_p$ are quite non-trivial. So, it is not easy, if not impossible, to invoke some convexity argument.)