# Sum of square roots inequality

For all $$a, b, c, d > 0$$, prove that $$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$

The idea would be to use AM-GM, but $$\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$ is hard to expand. I also tried squaring both sides, but that hasn't worked either. Using two terms at a time doesn't really work as well. How can I solve this question? Any help is appreciated.

You can use QM-AM for $$\sqrt{a},\ \sqrt{b},\ \sqrt{c},\ \sqrt{d}$$; I'm leaving the details for you.

If you square both sides:

$$(\sqrt a + \sqrt b + \sqrt c +\sqrt d)^2 =a + b+c+d + 2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})$$

while $$(2\sqrt {a+b+c+d})^2= 4(a+b+c+d)$$ so it suffices to prove

$$2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})\le 3(a+b+c+d)$$.

If we apply AM/GM though, we get $$\sqrt{ab} \le \frac {a+b}2$$ or in other words $$2\sqrt{ab} \le a+b$$.

And that does it:

$$2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})\le$$

$$(a+b) + (a+c) + (a+d) + (b+c) + (b+d) = 3(a+b+c+d)$$.

• Nice proof, I like it. Commented Jan 29, 2021 at 21:39

How about Jensen's inequality? $$\sqrt x$$ is concave so

$$\sqrt{\frac{a+b+c+d}{4}} \ge \frac 14 \left( \sqrt a +\sqrt b +\sqrt c +\sqrt d \right)$$

Alternative solution using Cauchy-Schwarz, which finishes the problem off immediately. By C-S, we have: \begin{align} & (a+b+c+d)(1+1+1+1) \geq (\sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d} )^2 \\ & \Rightarrow 4(a+b+c+d) \geq (\sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d} )^2 \\ & \Rightarrow 2\sqrt{a+b+c+d} \geq \sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d}. \end{align}