1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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