Suppose, for positive reals $a$, $b$, $c$, that $$ab+bc+ca+abc=4$$ Prove that $$\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq3$$

I applied AM-GM on the first equality ie, $a$, $b$, $c$, and $abc$ to get $$ab+bc+ca \geq 3\qquad\text{and}\qquad abc \leq1$$

The exact equation is as follows

$$1=\frac{ab+bc+ca+abc}{4}\ge\sqrt[4]{(abc)^3}\implies 1\ge abc$$

However, I didn't manage to get any further than this after applying AM-GM to several other inequalities.

I'd like a solution for this that utilizes AM-GM only, as I'm very new to inequalities.

  • $\begingroup$ Something is missing. If $a=b=c=-2$ the assumption $ab+bc+ca+abc=4$ holds, but the sum of those square roots is $6>3$. Did you forget to include the assumption that $a,b,c$ are positive? Exactly how did you apply AM-GM to the equation? Or, is something else missing given that you refer to the first inequality. $\endgroup$ – Jyrki Lahtonen Oct 15 '17 at 5:52
  • $\begingroup$ i forgot to write abc are positive reals and i meant first equality. $\endgroup$ – John Doe 1234 Oct 15 '17 at 6:00
  • $\begingroup$ If the condition is $ab+bc+ca=\mbox{const}$, the function $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}$ has a maximum. If the condition is $abc=\mbox{const}$, the function has a minimum. How do AM-GM equations know at what combination coefficients the minimum-maximum sign changes? $\endgroup$ – Zhuoran He Oct 15 '17 at 6:12
  • 1
    $\begingroup$ It's the same as this math.stackexchange.com/questions/1180443/…, just the condition is written differently. $\endgroup$ – Sil Oct 15 '17 at 6:12
  • $\begingroup$ For contest math that means a completely different trick is required. As in my point, even a change of the coefficients would mean a completely different problem. $\endgroup$ – Zhuoran He Oct 15 '17 at 6:14

Let $a=\frac{2x}{y+z}$ and $b=\frac{2y}{x+z}$, where $x$, $y$ and $z$ be positives.

Thus, the condition gives $$\frac{4xy}{(x+z)(y+z)}+2c\left(\frac{x}{y+z}+\frac{y}{x+z}\right)+\frac{4xyc}{(x+z)(y+z)}=4$$ or $$\frac{2c(x^2+y^2+xz+yz+2xy)}{(x+z)(y+z)}=4-\frac{4xy}{(x+z)(y+z)}$$ or $$\frac{2c(x+y)(x+y+z)}{(x+z)(y+z)}=\frac{4z(x+y+z)}{(x+z)(y+z)}$$ or $$c=\frac{2z}{x+y}$$ and we need to prove that $$2\sum_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\leq3, $$ which is AM-GM: $$2\sum_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\leq\sum_{cyc}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=$$ $$=\sum_{cyc}\left(\frac{x}{x+z}+\frac{z}{z+x}\right)=3. $$ Done!

There are proofs by trigonometry and $uvw$ but they are not easy.

  • $\begingroup$ How can you assume them as sharing the same variable...... $\endgroup$ – John Doe 1234 Oct 15 '17 at 7:19
  • $\begingroup$ That's the trick. The constraint $ab+bc+ac+abc=4$ is equivalent to $a=2x/(y+z)$, $b=2y/(x+z)$, $c=2z/(x+y)$ for arbitrary $x,y,z\in\mathbb{R}^+$. One degree of freedom is lost because rescaling $x,y,z$ by the same factor does not affect $a,b,c$. So the trick does not generalize to e.g. $2(ab+bc+ac)+abc=7$? $\endgroup$ – Zhuoran He Oct 15 '17 at 8:16
  • $\begingroup$ @Zhuoran Yes it does not work for your condition. But we can use another methods. We can use the contradiction method, for example. $\endgroup$ – Michael Rozenberg Oct 15 '17 at 8:30
  • $\begingroup$ I dont get it is there a simpler method. $\endgroup$ – John Doe 1234 Oct 15 '17 at 9:21
  • $\begingroup$ @John Doe 1234 What is your question. I am ready to explain. All other methods, which I see they are harder. $\endgroup$ – Michael Rozenberg Oct 15 '17 at 10:22

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