Prove: $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq\sqrt{x}+\sqrt{y}$

Question

$$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq\sqrt{x}+\sqrt{y}$$ where $x,y>0$

How should I approach this?

Answer 1

Fleshing out a comment I left below the OP, I find it convenient to recast problems like this in a form that gets rid of the square root symbol. In this case that's easily done by setting $\sqrt x=u$ and $\sqrt y=v$, which turns the inequality into

$${u^2\over v}+{v^2\over u}\ge u+v$$

An important thing to note is that we have $u,v\gt0$, since each variable stands for a square root (and appears in a denominator). That means we can argue as follows:

$$\begin{align} {u^2\over v}+{v^2\over u}\ge u+v&\iff u^3+v^3\ge u^2v+v^2u\\ &\iff u^2(u-v)+v^2(v-u)\ge 0\\ &\iff (u^2-v^2)(u-v)\ge0\\ &\iff (u+v)(u-v)^2\ge0 \end{align}$$

with the final line being clearly true since $u,v\gt0$ makes $u+v\ge0$, while $(u-v)^2\ge0$ even if the variables are allowed to be negative.

The basic argument here is no different, really, than in some of the other answers. I just find it easier to see what's going on, and avoid making mistakes, if I don't have a bunch of square root symbols to keep track of.

Answer 2

multiplying with $$\sqrt{xy}$$ we get $$x\sqrt{x}+y\sqrt{y}\geq x\sqrt{y}+y\sqrt{x}$$ and this is equivalent to $$(\sqrt{x}-\sqrt{y})(x-y)\geq 0$$ and this is true for $$x>0,y>0$$

Answer 3

You can try using AM-GM for $x,y>0$. $$ \sqrt{x} = \frac{\sqrt{x}}{y^{1/4}}\cdot y^{1/4} \leq \frac{1}{2}\left( \frac{x}{\sqrt{y}} + \sqrt{y}\right) $$ and, in a similar way, $$ \sqrt{y}\leq \frac{1}{2}\left( \frac{y}{\sqrt{x}} + \sqrt{x}\right). $$ Summing up the two inequalities you get the thesis.

Answer 4

Multiplying by $\sqrt{xy}$ (You can do this since $\sqrt{xy}>0$) you get $$x\sqrt{x}+y\sqrt{y}\geq x\sqrt{y}+y\sqrt{x}\\x(\sqrt{x}-\sqrt{y})\geq y(\sqrt{x}-\sqrt{y})\\(x-y)(\sqrt{x}-\sqrt{y})\geq 0$$ Since we have that $x,y\geq0$ we are left with two cases $x\geq y$ which implies $\sqrt x\geq \sqrt y$

And $x< y$ which implies $\sqrt{x}<\sqrt{y}$ both cases are almost trivially true.

Answer 5

Write $\sqrt{x}=\frac{x}{\sqrt{x}}$ and similarly for $y$ and you get:

$$\begin{align}\left(\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\right)-\left(\sqrt{x}+\sqrt{y}\right) &=x\left(\frac{1}{\sqrt{y}}-\frac{1}{\sqrt{x}}\right)+y\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}\right)\\ (x-y)\left(\frac{1}{\sqrt{y}}-\frac{1}{\sqrt{x}}\right)\\ &=\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{xy}} \end{align}$$

You get, more generally, if $f$ is an increasing function on the positive reals, then:

$$\frac{x}{f(y)}+\frac{y}{f(x)}\geq \frac{x}{f(x)}+\frac{y}{f(y)}$$

Answer 6

By C-S $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq\frac{(\sqrt{x}+\sqrt{y})^2}{\sqrt{y}+\sqrt{x}}=\sqrt{x}+\sqrt{y}$$

Answer 7

Multiply by $\sqrt{xy}$ both sides and with little algebraic manipulations you will get $(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})^2) \ge 0$ which is obvious.

Answer 8

One more way, this follows directly from Rearrangement inequality...

$$\frac{x}{\sqrt y} + \frac{y}{\sqrt x} \geqslant \frac{x}{\sqrt x} + \frac{y}{\sqrt y}$$

Answer 9

\begin{align} \frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}&=\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{xy}}\\ &=\frac{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})^2}{\sqrt{xy}}\ge 0. \end{align}

Answer 10

A detailed way of proving this.

We start by multiplying through by $\sqrt{x} \sqrt{y}$ $ $ , which upon simplification, gives $$x\sqrt{x}+y\sqrt{y}\geq x\sqrt{y}+y\sqrt{x}.$$

So, $$\Leftrightarrow x\sqrt{x}+y\sqrt{y} -x\sqrt{y}-y\sqrt{x} \geq 0$$ $$\Leftrightarrow x\sqrt{x}-x\sqrt{y}+y\sqrt{y}-y\sqrt{x} \geq 0$$ $$\Leftrightarrow x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x}) \geq 0.$$

But we don't know whether this last inequality holds or not (because it was derived from assuming your initial inequality was true). $ $ We can show this last inequality indeed holds as follows.

First of all, notice that the brackets always have the same magnitude, but can have different signs. That is, $$|\sqrt{x}-\sqrt{y}|=|\sqrt{y}-\sqrt{x}|.$$

Let's assume $ $ $y> x$ $ $.

(For simplicity's sake, which won't change the validity of the proof, as you'll see.)

Then it's true that $ $ $(\sqrt{y}-\sqrt{x}) > 0$ and $(\sqrt{x}-\sqrt{y}) < 0$.

This means that in $ $ $x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x}) \geq 0$, $ $ the factors $x$ and $y$ are both multiplied by a number of the same magnitude.

Therefore, it follows that the magnitude of the first term $x(\sqrt{x}-\sqrt{y})$ is smaller than the magnitude of the second term $y(\sqrt{y}-\sqrt{x})$. $ $ That is,

$$|x(\sqrt{x}-\sqrt{y})|<|y(\sqrt{y}-\sqrt{x})|.$$

Then, since the term $ $ $y(\sqrt{y}-\sqrt{x})$ is positive (because we assume $ $ $y> x$), it follows that $$x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x}) > 0.$$

If $y=x$, then it's clear that $$x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x}) = 0.$$ So, as required, it is true that $$x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x}) \geq 0.$$

(Note that if we assume $x>y$ instead of $y>x$, then analogously, the above also holds.)