It is equivalent to $$ \sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge\frac{a+b}{2}-\frac{2ab}{a+b} $$or $$ \frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+2\sqrt{ab}}\ge\frac{(a-b)^2}{2(a+b)} $$or $$ (a-b)^2(2(a+b)-2\sqrt{ab}-\sqrt{2(a^2+b^2)})\ge 0. $$Now, note that $$ \begin{aligned} 2(a+b)-2\sqrt{ab}-\sqrt{2(a^2+b^2)} &=(\sqrt{a}-\sqrt{b})^2-\frac{(a-b)^2}{a+b+\sqrt{2(a^2+b^2)}}= \\ &=\frac{(\sqrt{a}-\sqrt{b})^2}{a+b+\sqrt{2(a^2+b^2)}}(a+b+\sqrt{2(a^2+b^2)}-(\sqrt{a}+\sqrt{b})^2)= \\ &=\frac{(\sqrt{a}-\sqrt{b})^2(\sqrt{2(a^2+b^2)}-2\sqrt{ab})}{a+b+\sqrt{2(a^2+b^2)}}= \\ &=\frac{2(\sqrt{a}-\sqrt{b})^2(a-b)^2}{(a+b+\sqrt{2(a^2+b^2))(\sqrt{2(a^2+b^2)}+2\sqrt{ab})}}. \end{aligned} $$Thus, our inequality is equivalent to the following one: $$ \frac{2(\sqrt{a}-\sqrt{b})^2(a-b)^4}{(a+b+\sqrt{2(a^2+b^2))(\sqrt{2(a^2+b^2)}+2\sqrt{ab})}}\ge 0, $$or to $$ \frac{(a-b)^6}{(\sqrt{a}+\sqrt{b})^2(a+b+\sqrt{2(a^2+b^2))(\sqrt{2(a^2+b^2)}+2\sqrt{ab})}}\ge 0, $$or to $(a-b)^6\ge 0$. Hence, the inequality holds and the equality holds if and only if $a=b$.

Absolutely straightforward when you use $s=a+b$, $p=ab$, with $s\geq 2\sqrt{p}$. We wish to prove $\sqrt{\frac{s^2-2p}{2}} + \frac{2p}{s} \geq \frac{s}{2} + \sqrt{p}$, i.e. $\sqrt{\frac{s^2-2p}{2}} - \sqrt{p} \geq \frac{s^2-4p}{2s}$. Both sides are non-negative since $s \geq 2\sqrt{p}$, so we can square them to obtain the equivalent $\frac{s^2}{2} - \sqrt{2p(s^2-2p)} \geq \frac{s^4 - 8s^2p + 16p^2}{4s^2}$, i.e. $\frac{s^4 + 8s^2p - 16p^2}{4s^2} \geq \sqrt{2p(s^2-2p)}$. After squaring again and opening the brackets we obtain $256 p^4 - 256 p^3 s^2 + 96 p^2 s^4 - 16 p s^6 + s^8 \geq 0$, equivalent to $(4p-s^2)^4 \geq 0$ which clearly holds. Equality holds if and only if $s^2-4p = (a-b)^2 = 0$, i.e. $a=b$.

StefanSebez wrote: Prove that for all positive real numbers $a$, $b$ the following inequality holds: \begin{align*} \sqrt{\frac{a^2+b^2}{2}}+\frac{2ab}{a+b}\ge \frac{a+b}{2}+ \sqrt{ab} \end{align*}When does equality hold?

Let $a,b$ be positive real numbers . Prove that$$\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}\geq\frac{a^2+b^2}{a+b}+\sqrt{ab}.$$

Let $a,b$ be positive real numbers . Prove that$$\sqrt[3]{\frac{a^3+b^3}{2}}+\frac{3ab}{a+b}\ge \frac{a+b+ 3\sqrt{ab}}{2}$$

Prove that for all positive real numbers $a$, $b$ the following inequality holds: \begin{align*} 3\sqrt{\frac{a^2+ab+b^2}{3}}+\frac{2ab}{a+b}\ge \frac{3(a+b)}{2}+ \sqrt{ab} \end{align*} StefanSebez wrote: Prove that for all positive real numbers $a$, $b$ the following inequality holds: \begin{align*} \sqrt{\frac{a^2+b^2}{2}}+\frac{2ab}{a+b}\ge \frac{a+b}{2}+ \sqrt{ab} \end{align*}When does equality hold? sqing wrote: Let $a,b$ be positive real numbers . Prove that$$\sqrt[3]{\frac{a^3+b^3}{2}}+\frac{3ab}{a+b}\ge \frac{a+b+ 3\sqrt{ab}}{2}$$

Let $a,b,c$ be positive real numbers . Prove that$$2\sqrt{\frac{a^2+b^2+c^2}{3}}+\frac{3abc}{ab+bc+ca}\ge \frac{2(a+b+c)+3 \sqrt[3]{abc}}{3}$$$$5\sqrt{\frac{a^2+b^2+c^2}{3}}+\frac{3abc}{a^2+b^2+c^2}\ge \frac{5(a+b+c)+3 \sqrt[3]{abc}}{3}$$

Since it is homogenous, we can let $a+b=2$ Then you can write it as a very manageable expression in terms of a single variable $x = ab$ After squaring once, the desired result becomes obviously true by AM / GM. PS - The whole expression can be written in terms of the Quadratic, Arithmetic, and Geometric Mean. Has anybody explored the significance of this?

StefanSebez wrote: Prove that for all positive real numbers $a$, $b$ the following inequality holds: \begin{align*} \sqrt{\frac{a^2+b^2}{2}}+\frac{2ab}{a+b}\ge \frac{a+b}{2}+ \sqrt{ab} \end{align*}When does equality hold? AM-GM on $(a+b)^4$ and $8ab\left(a^2+b^2\right)$ $\frac{(a+b)^4+8ab\left(a^2+b^2\right)}{2}\ge\sqrt{(a+b)^4(8ab)\left(a^2+b^2\right)}$ $\frac{(a+b)^4+8ab\left(a^2+b^2\right)}{2}\ge2(a+b)^2\sqrt{2ab\left(a^2+b^2\right)}$ $\frac{(a+b)^4+8ab\left(a^2+b^2\right)}{4(a+b)^2}\ge\sqrt{2ab\left(a^2+b^2\right)}$ $\frac{(a+b)^4+(4ab)\left(2a^2+2b^2\right)}{4(a+b)^2}\ge\sqrt{2ab\left(a^2+b^2\right)}$ $\frac{(a+b)^4}{4(a+b)^2}+\frac{\left((a+b)^2-(a-b)^2\right)\left((a+b)^2+(a-b)^2\right)}{4(a+b)^2}\ge\sqrt{2ab\left(a^2+b^2\right)}$ $\frac{(a+b)^4+(a+b)^4-(a-b)^4}{4(a+b)^2}\ge\sqrt{2ab\left(a^2+b^2\right)}$ $\frac{2(a+b)^4-(a-b)^4}{4(a+b)^2}\ge\sqrt{2ab\left(a^2+b^2\right)}$ $\frac{2(a+b)^4}{4(a+b)^2}-\frac{(a-b)^4}{4(a+b)^2}\ge\sqrt{2ab\left(a^2+b^2\right)}$ $\frac{(a+b)^2}{2}-\frac{(a-b)^4}{4(a+b)^2}\ge\sqrt{2ab\left(a^2+b^2\right)}$ $\frac{(a+b)^2}{2}-\sqrt{2ab\left(a^2+b^2\right)}\ge\frac{(a-b)^4}{4(a+b)^2}$ $\frac{a^2+b^2}{2}-2\sqrt{ab}\sqrt{\frac{a^2+b^2}{2}}+ab\ge\frac{(a-b)^4}{4(a+b)^2}$ $\left(\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\right)^2\ge\left(\frac{(a-b)^4}{2(a+b)}\right)^2$ $\left(\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\right)^2\ge\left(\frac{a^2-2ab+b^2}{2(a+b)}\right)^2$ $\left(\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\right)^2\ge\left(\frac{(a+b)^2-4ab}{2(a+b)}\right)^2$ $\left(\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\right)\ge\left(\frac{a+b}{2}-\frac{2ab}{a+b}\right)^2$ $\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge\frac{a+b}{2}-\frac{2ab}{a+b}$ $\sqrt{\frac{a^2+b^2}{2}}+\frac{2ab}{a+b}\ge\frac{a+b}{2}+\sqrt{ab}$

richrow12 wrote: It is equivalent to $$ \sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge\frac{a+b}{2}-\frac{2ab}{a+b} $$or $$ \frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+2\sqrt{ab}}\ge\frac{(a-b)^2}{2(a+b)} $$or $$ (a-b)^2(2(a+b)-2\sqrt{ab}-\sqrt{2(a^2+b^2)})\ge 0. $$Now, note that $$ \begin{aligned} 2(a+b)-2\sqrt{ab}-\sqrt{2(a^2+b^2)} &=(\sqrt{a}-\sqrt{b})^2-\frac{(a-b)^2}{a+b+\sqrt{2(a^2+b^2)}}= \\ &=\frac{(\sqrt{a}-\sqrt{b})^2}{a+b+\sqrt{2(a^2+b^2)}}(a+b+\sqrt{2(a^2+b^2)}-(\sqrt{a}+\sqrt{b})^2)= \\ &=\frac{(\sqrt{a}-\sqrt{b})^2(\sqrt{2(a^2+b^2)}-2\sqrt{ab})}{a+b+\sqrt{2(a^2+b^2)}}= \\ &=\frac{2(\sqrt{a}-\sqrt{b})^2(a-b)^2}{(a+b+\sqrt{2(a^2+b^2))(\sqrt{2(a^2+b^2)}+2\sqrt{ab})}}. \end{aligned} $$Thus, our inequality is equivalent to the following one: $$ \frac{2(\sqrt{a}-\sqrt{b})^2(a-b)^4}{(a+b+\sqrt{2(a^2+b^2))(\sqrt{2(a^2+b^2)}+2\sqrt{ab})}}\ge 0, $$or to $$ \frac{(a-b)^6}{(\sqrt{a}+\sqrt{b})^2(a+b+\sqrt{2(a^2+b^2))(\sqrt{2(a^2+b^2)}+2\sqrt{ab})}}\ge 0, $$or to $(a-b)^6\ge 0$. Hence, the inequality holds and the equality holds if and only if $a=b$. I think the easiest solution is by $$ \frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+2\sqrt{ab}}\ge\frac{(a-b)^2}{2(a+b)} $$just square the both equations and you will get $$\frac{{a^2+b^2+2ab}}{2}\ge \sqrt{2ab(a^2+b^2)}$$which is just simple AM GM inequality.

Let $a=x+y,b=x-y, x >y \geq 0$ $\sqrt{x^2+y^2}+\frac{(x^2-y^2)}{x} \geq x+\sqrt{x^2-y^2}$ or $\sqrt{x^2+y^2} -\sqrt{x^2-y^2}\geq \frac{y^2}{x}$ or $\frac{2y^2}{\sqrt{x^2+y^2} +\sqrt{x^2-y^2}} \geq \frac{y^2}{x}$ or $x \geq \frac{\sqrt{x^2+y^2}+\sqrt{x^2-y^2}}{2}$ which is QM-AM