Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}$$

It seems nice enough.

I proved this inequality by Holder, but it quits very ugly.

Maybe there is something nice? Thank you!

  • $\begingroup$ Can this inequality be useful?$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$....See if this can be used anyway.... $\endgroup$ – tatan Sep 30 '16 at 6:41
  • $\begingroup$ No! Nesbitt is very weak here. $\endgroup$ – Michael Rozenberg Sep 30 '16 at 6:51
  • $\begingroup$ Isn't $a=b=c=1$ the solution that maximizes the RHS and minimizes the LHS? $\endgroup$ – Alex Silva Sep 30 '16 at 11:48
  • 1
    $\begingroup$ @AlexSilva No, that is not the case. The maximum of the RHS is $ 3 / \sqrt{2} $, which does indeed occur when $a = b = c = 1$. However, the LHS can be smaller. Take for instance $a = b = x$ and $c = 1/x^2$. Then the LHS is $2\sqrt{x^3 / (x^3 + 1)} + \sqrt{1/(2x^3)}$, which tends to $2$ as $x \to \infty$. And $2 < 3/\sqrt{2}$. $\endgroup$ – Dylan Oct 2 '16 at 0:35
  • 1
    $\begingroup$ Concerning the LHS, 2 is indeed the smallest value one can get. There is a general proof that, for $0 < k < 1$, one has $(\frac{a}{b+c})^k+(\frac{b}{c+a})^k+(\frac{c}{a+b})^k\geq$ min $(2 ; \frac{3}{2^k})$ given in Pham Kim Hung's book "Secrets in Inequalities (volume 2)", pp. 284 ff. $\endgroup$ – Andreas Oct 2 '16 at 15:34

Put the following substitution :


Remark that :


And :

$$abc=1 \iff -2=-\frac{1}{xyz}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

So we get the inequality related to this Post that you have proved by your own with brio .


Making $c=\frac{1 } {ab}$ the expression becomes $$f(a,b)= \left(\sqrt{\frac{a^2b}{ab^2+1}}+\sqrt{\frac{ab^2}{a^2b+1}}+\sqrt{\frac{1}{ab(a+b)}}\right)\sqrt{\frac{ab(a+b+15)+1}{ab}}\ge9$$ for all positive $a,b$.

It follows $$\sqrt{\frac{a^3b+a^2b^2+15a^2b+a}{ab^2+1}}+\sqrt{\frac{ab^3+a^2b^2+15ab^2+b}{a^2b+1}}+\frac{1}{ab}\sqrt{\frac{ab(a+b+15)+1}{a+b}} \ge9$$ It is clear $f(x,y)$ has no maximum and, in order to prove the inequality, we want to get the minimum of $f(x,y)$.

This minimum can be calculated as usually for two variables ($f_x(x,y)=0$ and $f_y(x,y)=0$, etc).

We calculate as follows: since $f(a,b)=f(b,a)$ the minimum of $f(a,b)$ is equal to the minimum of $f(a,a)$ where $a\gt 0$. Hence we calculate the minimum of the function of one variable $$f(x,x)=2\sqrt{\frac{2x^4+15x^3+x}{x^3+1}}+\frac{1}{x^2}\sqrt{\frac{2x^3+15x^2+1}{2x}}$$

The calculation is straightforward although somewhat tedious giving the minimum $9$ for $x=1$. For further explanation, see figure below wherein the calculation (Wolfram) and the graph of the function (Desmos) confirm the result. Thus this minimum is attained with $a = b = c = 1$ and the proposed inequality is valid for all positive with $abc=1$.

enter image description here

  • $\begingroup$ You are welcome to show us your calculations with derivatives. $\endgroup$ – Michael Rozenberg Oct 5 '16 at 17:07

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.