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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.

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  • $\begingroup$ Are you open to proofs that don't use AM-GM? @DrZafarAhmedDSc's answer presents an alternative. $\endgroup$ – J.G. Jul 17 at 6:59
  • $\begingroup$ 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. $\endgroup$ – AYA Jul 17 at 7:13
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Note that for $x,y \gt 0$, we have$\sqrt{x+y}=\sqrt {\sqrt{x}^2+\sqrt y^2}\lt \sqrt{\sqrt x^2+\sqrt y^2+2\sqrt x\sqrt y}=\sqrt{(\sqrt x+\sqrt y) ^2} =\sqrt x+ \sqrt y$

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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.

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    $\begingroup$ 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. $\endgroup$ – paul blart math cop Jul 17 at 7:00
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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.

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    $\begingroup$ 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. $\endgroup$ – Batominovski Jul 18 at 20:17
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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: $$\frac{a^2d^2+b^2c^2}{2}=\frac{|ad|^2+|bc|^2}{2}\geq \sqrt{|ad|^2\cdot |bc|^2}=|a|\,|b|\,|c|\,|d|\,.$$

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