# How to prove that if $x>0$ and $y>0$, then $\sqrt{x}+\sqrt{y}>\sqrt{x+y}$, using the relation of arithmetic and geometric means?

How to prove that if $$x>0$$ and $$y>0$$, then $$\sqrt{x}+\sqrt{y}>\sqrt{x+y}\,,$$ using the relation of arithmetic and geometric means?

I started by showing that if $$x>0$$ and $$y>0$$, based on the relation of arithmetic and geometric means, $$\dfrac{x+y}{2}\ge\sqrt{xy}$$.

Hence, $$x+y\ge2\sqrt{xy}$$.

I am now stuck here and don't know what must be the next step.

Any suggestions or comments will be much appreciated.

• Are you open to proofs that don't use AM-GM? @DrZafarAhmedDSc's answer presents an alternative. – J.G. Jul 17 at 6:59
• Thank you but I would like to know how I can use the relationship of AM-GM to prove this. I was able to prove this without the use of said relationship. My professor says that we can use AM-GM. – AYA Jul 17 at 7:13

Note that for $$x,y \gt 0$$, we have$$\sqrt{x+y}=\sqrt {\sqrt{x}^2+\sqrt y^2}\lt \sqrt{\sqrt x^2+\sqrt y^2+2\sqrt x\sqrt y}=\sqrt{(\sqrt x+\sqrt y) ^2} =\sqrt x+ \sqrt y$$

If $$x,y>0$$, then squaring the following on both sides $$\sqrt{x}+\sqrt{y}>\sqrt{x+y}$$ we have, $$x+y+2\sqrt{xy} > x+y \implies 2\sqrt{xy} >0,$$ which is true.

• It would be better to write these implications as if and only ifs (which is just as easily true). Showing that $P$ implies a true statement does not prove that it is true. – paul blart math cop Jul 17 at 7:00

Is this correct?

Since $$x+y≥2\sqrt{xy}$$, by AM-GM relationship,

$$x+y+2\sqrt{xy}≥x+y$$

$$(\sqrt{x}+\sqrt{y})^2≥x+y$$

Thus, $$\sqrt{x}+\sqrt{y}>\sqrt{x+y}$$. QED.

• While the inequality $x+y+2\sqrt{xy}\geq x+y$ is true, I don't see how you concluded from the AM-GM Inequality you wrote. – Batominovski Jul 18 at 20:17

Theorem (A Baby Version of the Triangle Inequality). Let $$a$$, $$b$$, $$c$$, and $$d$$ be real numbers. We have $$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+c)^2+(b+d)^2}\,.$$ The equality holds if and only if

• $$(a,b)=(0,0)$$, or
• there exists $$\lambda\geq 0$$ such that $$(c,d)=(\lambda a,\lambda b)$$.

Let $$x$$ and $$y$$ be nonnegative real numbers. Note that $$\sqrt{x}+\sqrt{y}=\sqrt{\sqrt{x}^2+0^2}+\sqrt{0^2+\sqrt{y}^2}\,.$$ By the Baby Triangle Inequality above, $$\sqrt{\sqrt{x}^2+0^2}+\sqrt{0^2+\sqrt{y}^2}\geq \sqrt{(\sqrt{x}+0)^2+(0+\sqrt{y})^2}\,.$$ That is, $$\sqrt{x}+\sqrt{y}\geq \sqrt{x+y}\,.\tag{#}$$ By the equality conditions of the Baby Triangle Inequality, (#) is an equality if and only if $$x=0$$ or $$y=0$$.

Now, I shall prove the Baby Triangle Inequality using the AM-GM Inequality to fulfill the OP's request that the AM-GM Inequality must be used. By squaring the required inequality, what we need to prove is equivalent to $$\sqrt{a^2+b^2}\sqrt{c^2+d^2}\geq ac+bd\,.$$ We, in fact, have a stronger inequality: $$\sqrt{a^2+b^2}\sqrt{c^2+d^2}\geq |a|\,|c|+|b|\,|d|\,. \tag{*}$$ By squaring the inequality above, we know that (*) is equivalent to $$a^2d^2+b^2c^2\geq 2\,|a|\,|b|\,|c|\,|d|\,,$$ which is true by the AM-GM Inequality: $$\frac{a^2d^2+b^2c^2}{2}=\frac{|ad|^2+|bc|^2}{2}\geq \sqrt{|ad|^2\cdot |bc|^2}=|a|\,|b|\,|c|\,|d|\,.$$