# Is this simple proof about symmetric sums correct?

I'm asked to prove $$x \gt 0, y \gt 0, z \gt 0 \rightarrow$$ $$\left(\frac{x+y}{x+y+z}\right)^\frac{1}{2}+\left(\frac{x+z}{x+y+z}\right)^\frac{1}{2} + \left(\frac{y+z}{x+y+z}\right)^\frac{1}{2} \le 6^\frac{1}{2}$$

I rewrite the summands and say that it is sufficient to prove:
$$A^\frac{1}{2} + B^\frac{1}{2} + C^\frac{1}{2} \le 6^\frac{1}{2}$$ $$A +B +C = 2$$ $$0 \lt A, B, C \le 1$$

Now I just square both sides to get:
$$A + (AB)^\frac{1}{2} + (AC)^\frac{1}{2} + B + (BC)^\frac{1}{2} + C \le 6$$

This seems simple: $$2 + (AB)^\frac{1}{2} + (AC)^\frac{1}{2} + (BC)^\frac{1}{2} \le$$ $$2 + 1 + 1 + 1 \le 5 \le 6$$

So I ended up proving that the original bounds were too loose. That makes me worry that I messed up my proof somewhere.

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Your expression for the square of $A^{1/2}+B^{1/2}+C^{1/2}$ is not correct. –  Andres Caicedo Feb 17 '13 at 17:25
You made a mistake when squaring. –  Ishan Banerjee Feb 17 '13 at 17:25
$(A^{\frac12}+B^{\frac12}+C^{\frac12})^2=A+B+C+2(\sqrt{AB}+\sqrt{AC}+\sqrt{BC})$ –  Hagen von Eitzen Feb 17 '13 at 17:29
Oh I see it now. I will have to think of a different proof then. –  Mark Feb 17 '13 at 17:31

You are almost there. Squaring both sides yields $$A+B+C+2(AB)^{1/2}+2(BC)^{1/2}+2(CA)^{1/2}\leq 6,$$ and so you only need to prove that $$(AB)^{1/2}+(BC)^{1/2}+(CA)^{1/2}\leq 2.$$ This inequality follows directly from the fact that $A+B+C=2$ along with Cauchy Schwarz.

Alternative Solution: Before squaring both sides, we can deduce desired inequality directly using Cauchy Schwarz. We have that $$\left(A^{1/2}+B^{1/2}+C^{1/2}\right)^2\leq (A+B+C)(1+1+1),$$ and so we see that $$A^{1/2}+B^{1/2}+C^{1/2}\leq 6^{1/2}.$$

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I understand your alternative solution, but can you elaborate on the use of Cauchy Schwarz in main solution? –  Mark Feb 17 '13 at 17:44
The vectors <A^.5, B^.5, C^.5> and <B^.5, C^.5, A^.5> will work to complete the proof. Thanks! –  Mark Feb 17 '13 at 19:11

Another approach is to use Lagrange multipliers. With $A$, $B$, and $C$ as you chose, the minimum value of $f(A,B,C) = \sqrt{A} + \sqrt{B} + \sqrt{C}$ subject to the constraint $A+B+C=2$ must occur at a local minimum of $F(A,B,C,\lambda) = \sqrt{A} + \sqrt{B} + \sqrt{C} + \lambda (A+B+C-2)$.

This can only occur if ${\partial{F}\over\partial{A}}$, ${\partial{F}\over\partial{B}}$, and ${\partial{F}\over\partial{C}}$ are simultaneously zero. Calculating the derivatives, we get $${1\over{2\sqrt{A}}}+\lambda={1\over{2\sqrt{B}}}+\lambda={1\over{2\sqrt{C}}}+\lambda=0\textrm{,}$$ which implies that $A=B=C$. Using the fact that $A+B+C=2$, it follows that $A=B=C={2\over3}$. The minimum value of $\sqrt{A} + \sqrt{B} + \sqrt{C}$ is then $3\sqrt{2\over3}$, which equals $\sqrt6$.

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