Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Your expression for the square of $A^{1/2}+B^{1/2}+C^{1/2}$ is not correct. – Andrés E. 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
up vote 1 down vote accepted

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 $$A^{1/2}+B^{1/2}+C^{1/2}\leq 6^{1/2}.$$

share|cite|improve this answer
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 $\sqrt6$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.