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Given that $x$ and $y$ are positives, find the minimum value of $$\large \sqrt{\frac{x^2}{y^2} + \frac{y^2}{x^2} + 2} + \frac{\sqrt{xy}}{x + y}$$

We have that $\sqrt{\dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} + 2} = \sqrt{\left(\dfrac{x}{y}\right)^2 + \left(\dfrac{y}{x}\right)^2 + 2 \cdot \dfrac{x}{y} \dfrac{y}{x}}$

$= \sqrt{\left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2} = \dfrac{x}{y} + \dfrac{y}{x} = \dfrac{x^2 + y^2}{xy}$. Furthermore, $2(x^2 + y^2) \ge (x + y)^2$

$\implies \dfrac{x^2 + y^2}{xy} \ge \dfrac{(x + y)^2}{2xy} \implies \sqrt{\dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} + 2} \ge \dfrac{(x + y)^2}{2xy}$.

Using the AM-GM inequality, we have that $$\frac{(x + y)^2}{2xy} + 2 \cdot \frac{\sqrt{xy}}{2(x + y)} \ge 3 \cdot \sqrt[3]{\frac{(x + y)^2}{2xy} \left[\frac{\sqrt{xy}}{2(x + y)}\right]^2} = \frac{3}{2}$$

$\implies \sqrt{\dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} + 2} + \dfrac{\sqrt{xy}}{x + y} \ge \dfrac{3}{2}$. But the equality sign happens when $x = y = 0$, which contradicts the fact that $x, y > 0$.

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  • $\begingroup$ Equality happens when $x=y$, not necessarily $=0$ $\endgroup$ – Hagen von Eitzen May 30 '19 at 10:35
  • $\begingroup$ No, if you solve all of the condition, you have $x = y$ and $x + y - \sqrt{xy} = 0$, which is only possible if $x = y = 0$. $\endgroup$ – Lê Thành Đạt May 30 '19 at 10:37
  • $\begingroup$ $\sqrt{\frac{x^2}{y^2}+\frac{y^2}{x^2}+2}\ge \sqrt{2+2}=2$. $\frac{3}{2}$ is a lower bound but cannot be reached. $\endgroup$ – CY Aries May 30 '19 at 10:46
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By AM-GM, C-S and AM-GM we obtain: $$\sqrt{\frac{x^2}{y^2}+\frac{y^2}{x^2}+2}+\frac{\sqrt{xy}}{x+y}=\frac{1}{2}\left(4\cdot\frac{x^2+y^2}{2xy}+\frac{2\sqrt{xy}}{x+y}\right)\geq$$ $$\geq\frac{5}{2}\sqrt[5]{\left(\frac{x^2+y^2}{2xy}\right)^4\frac{2\sqrt{xy}}{x+y}}=\frac{5}{2}\sqrt[5]{\frac{(x^2+y^2)^4}{8x^3y^3\sqrt{xy}(x+y)}}\geq\frac{5}{2}\sqrt[5]{\frac{(x^2+y^2)^3\cdot\frac{(x+y)^2}{2}}{8x^3y^3\sqrt{xy}(x+y)}}\geq\frac{5}{2}.$$ The equality occurs for $x=y,$ which says that we got a minimal value.

Another way.

Let $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=2k.$

Thus, $k\geq1$ and we need to prove that: $$4k^2-2+\frac{1}{2k}\geq\frac{5}{2}$$ or $$(k-1)(8k^2+8k-1)\geq0,$$ which is obvious.

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Let $t=\sqrt{\frac xy}>0$. Then $$ \begin{align}\sqrt{\frac{x^2}{y^2}+\frac{y^2}{x^2}+2}+\frac{\sqrt{xy}}{x+y}&=\sqrt{t^4+t^{-4}+2}+\frac1{t+t^{-1}}\\ &=\sqrt{(t^2+t^{-2})^2}+\frac1{t+t^{-1}}\\ &=t^2+t^{-2}+\frac1{t+t^{-1}}\\ &=(t+t^{-1})^2-2+\frac1{t+t^{-1}}\\ &=u^2+u^{-1}-2\end{align}$$ with $u:=t+t^{-1}$. Note that $$ u=t-2+t^{-1}+2=\left(\sqrt t-\frac1{\sqrt t}\right)^2+2\ge 2$$ with equality if and only if $t=1$ (i.e., if and only if $x=y$). Next, $u\mapsto u^2+u^{-1}$ is strictly increasing$^1$ on $[2,\infty)$ and we conclude $$\sqrt{\frac{x^2}{y^2}+\frac{y^2}{x^2}+2}+\frac{\sqrt{xy}}{x+y}\ge 2^2+2^{-1}-2=\frac52$$ with equality if and only if $x=y$.


$^1$ This follows either by looking at the derivative $2u-u^{-2}\ge 4-\frac 14$. But in the proof we could alternatively forget about derivatives and use $$u^2+u^{-1}-2=\left(\frac{\sqrt u}2-\frac1{\sqrt u}\right)^2 -1+u\cdot \left(u-\frac14\right)\ge 0-1+2\cdot \frac{7}4=\frac 52.$$

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We can all agree that $\sqrt{\dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} + 2} \ge \dfrac{(x + y)^2}{2xy}$. What's next is different.

We have that $\dfrac{(x + y)^2}{2xy} + \dfrac{\sqrt{xy}}{x + y} = \dfrac{7(x + y)^2}{16xy} + \left[\dfrac{(x + y)^2}{16xy} + 2 \cdot \dfrac{\sqrt{xy}}{2(x + y)}\right]$

Using the AM-GM inequality, it is evident that $$\frac{(x + y)^2}{16xy} + 2 \cdot \frac{\sqrt{xy}}{2(x + y)} \ge 3 \cdot \sqrt[3]{\frac{(x + y)^2}{16xy}\left[\dfrac{\sqrt{xy}}{2(x + y)}\right]^2} = \dfrac{3}{4}$$

In addition, $(x + y)^2 \ge 4xy \implies \dfrac{7(x + y)^2}{16xy} \ge \dfrac{7}{4}$.

To sum it up, $\sqrt{\dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} + 2} + \dfrac{\sqrt{xy}}{x + y} \ge \dfrac{3}{4} + \dfrac{7}{4} = \dfrac{5}{2}$.

The equality sign happens when $\left\{ \begin{align*} (x + y)^2 = 2(x^2 + y^2)\\ \dfrac{(x + y)^2}{16xy} = \dfrac{\sqrt{xy}}{x + y}\\ (x + y)^2 = 4xy \end{align*}\\ \right.$$\implies x = y$.

Are there any other solutions that are more practical?

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Assume

$$xy= 1\,\therefore\,{y}'= -\,\frac{y}{x}$$

$$\therefore\,(\,x- y\,)(\,x- 1\,)\geqq 0$$

$$\because\,x\geqq y \tag{ASSUME!ing}$$

We let

$$w(\,x\,)= \frac{x}{y}+ \frac{y}{x}+ \frac{\sqrt{xy}}{x+ y}- \frac{5}{2}= (\,x+ y\,)^{\,2}+ \frac{1}{x+ y}- \frac{9}{2}$$

$$\therefore\,{w}'(\,x\,)= \frac{(\,{y}'+ 1\,)[\,2(\,x+ y\,)^{\,3}- 1]}{(\,x+ y\,)^{\,2}}$$

$$\therefore\,{w}'(\,x\,)= \frac{(\,x- y\,)[\,2(\,x+ y\,)^{\,3}- 1]}{x(\,x+ y\,)^{\,2}}$$

Using above that I found

$$\therefore\,(\,x- 1\,){w}'(\,x\,)\geqq 0$$

$$\therefore\,w(\,x\,)\geqq w(\,1\,)= 0$$

$$\because\,(\,x- 1\,){w}'(\,x\,)= w(\,x\,)- w(\,1\,) \tag{tangent equation}$$

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