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Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive

$$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$

$$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$

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$$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$

$$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$

$$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$

$$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$

$$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$

All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$

Is the proof valid?

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If $x$ and $y$ are assumed to be positive, then I do not see anything wrong, but I think it would be *much* better if you rewrote it from the bottom up instead of as you have it now (working *from* what you are trying to prove as opposed to *to* it, if that makes sense).

Alternatively, apply AM-GM inequality with $a = \sqrt{x}, b = \sqrt{y}$:

$\dfrac{a^2}{b} + b \geq 2\sqrt{\dfrac{a^2}{b}\cdot b} = 2a, \dfrac{b^2}{a} + a \geq 2\sqrt{\dfrac{b^2}{a}\cdot a} = 2b$. Adding these $2$ inequalities to get the answer.

**Generalization**: The same technique could be used to prove: If $x, y, z > 0 \implies \dfrac{x^2}{y} + \dfrac{y^2}{z} + \dfrac{z^2}{x} \geq x+y+z$.

First you neeed to qualify your starting point by saying that $x,y > 0$ and that by $\sqrt{x}$ or $\sqrt{y}$ the proposition means the positive value of that square root. But these are tirival details.

The second flaw is that what you presented is not a proof; the very first line of the “proof” asserts the proposition that was to have been proven. The steps you present are a very good way to discover the steps of the proof, but the proof itself would have to start with your last step and work its way upwards. Thus:

$$(\sqrt{x}-\sqrt{y})^2 \geq 0 \wedge (\sqrt{x}+\sqrt{y}) \geq 0 \wedge \sqrt{y}\sqrt{x} > 0 \implies

\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$

should have been the first step, not the last. From there you go to your next-to-last step, and onward till you have reached the proposition to be proven.

Use Titus lemma:

$$\frac{x^2}{a}+\frac{y^2}b\ge \frac{(x+y)^2}{a+b}$$

Then $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}=\frac{\sqrt{x}^2}{\sqrt{y}}+\frac{\sqrt{y}^2}{\sqrt{x}}\ge \frac{(\sqrt{x}+\sqrt{y})^2}{(\sqrt{x}+\sqrt{y})}=\sqrt{x}+\sqrt{y}$$

Your proof is valid, but I suggest an alternative.

Let $a=\sqrt{x}>0$ and $b=\sqrt{y}>0$. You want to show

$$

\frac{a^2}{b}+\frac{b^2}{a}\ge a+b

$$

which becomes

$$

\frac{a^3+b^3}{ab}\ge a+b

$$

or

$$

\frac{(a+b)(a^2-ab+b^2)}{ab}\ge a+b

$$

Since $a+b>0$ and $ab>0$, we can factor out $a+b$ and remove the denominator, so the inequality becomes

$$

a^2-ab+b^2\ge ab

$$

Can you finish?

Your strategy works as well: the inequality is equivalent to

$$

\frac{a^2}{b}+\frac{b^2}{a}-a-b\ge0

$$

that becomes

$$

\frac{a^3+b^3-ab(a+b)}{ab}\ge0

$$

The numerator can be rewritten as

$$

a^3+b^3-ab(a+b)=(a+b)(a^2-ab+b^2)-ab(a+b)=

(a+b)(a-b)^2

$$

So the inequality is

$$

\frac{(a+b)(a-b)^2}{ab}\ge0

$$

which is true.

There is a simpler way, by eliminating at once the square roots.

Setting $x=X^2$ and $y=Y^2$ with $X>0$ and $Y>0$, the issue is equivalent to:

$$\text{show that} \ \ \ \dfrac{X^2}{Y}+\dfrac{Y^2}{X} \geq X+Y$$

which amounts to prove that:

$$X^3+Y^3 \geq XY(X+Y)$$

knowing factorization $X^3+Y^3=(X+Y)(X^2-XY+Y^2)$ and simplifying by $X+Y$, one is brought back to show that $X^2-2XY+Y^2 \geq 0$ i.e. $(X-Y)^2\geq 0$ which is eviedently true.

There are three case you can show to prove this to be true. The first case is when x=y. Here you can just substitute x in for y. Therefore, the inequality becomes $\frac{x}{\sqrt x} + \frac{x}{\sqrt x} \ge {\sqrt x} + {\sqrt x}$. We can simplify this to $2x^\frac{3}{2} \ge 2{\sqrt x}$. This is clearly true. Now we can show that this is true for some x greater than y. For this to to be true x = ay, where a is some constant greater than 1. Therefore we plug this into the inequality. and find $\frac{ay}{\sqrt y} + \frac{y}{\sqrt ay} \ge {\sqrt ay} + {\sqrt y}$. Simplifying the left hand side, we find $\frac{a(y)^\frac{3}{2}+y^\frac{3}{2}}{(\sqrt a)(y)} \ge {\sqrt ay} + {\sqrt y}$. Multiplying the denominator over, we get the left hand side to be $ (a)(y^\frac{3}{2})+(\sqrt a)(y^\frac{3}{2})$. This leaves our inequality as $(a)(y)^\frac{3}{2}+y^\frac{3}{2} \le (a)(y^\frac{3}{2})+(\sqrt a)(y^\frac{3}{2})$. The same can be shown for y = ax.

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