Let $a,b,c>0$ satisfy $abc=1$, prove that: $$\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$$ My attempt: Let $a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}$, we have $xyz=1$ and using $abc=1$, the inequality can be written as: $$\dfrac{x}{\sqrt{x+y}}+\dfrac{y}{\sqrt{y+z}}+\dfrac{z}{\sqrt{z+x}}\le \dfrac{xy+yz+zx}{\sqrt{2}}$$ I'm trying to use Cauchy-Schwarz: $$LHS\le\sqrt{(x+y+z)(\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x})}$$ but now I have to prove $$\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}\le\dfrac{3}{2}$$ because $ab+bc+ca\ge\sqrt{3(a+b+c)}$, but I can't prove it. Can anyone give me a hint? Not necessarily a complete solution.

By the way, I also relized a problem that seems quite similar to the above problem $\sqrt{\frac{2 x}{x+y}}+\sqrt{\frac{2 y}{y+z}}+\sqrt{\frac{2 z}{z+x}} \leq 3$ if $x,y,z>0$ (Vasile Cirtoaje) (and then we can use $3\le xy+yz+zx$ ?Hope it helps)

  • 1
    $\begingroup$ $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}\le\dfrac{3}{2}$ is not true. Please check $x = 2, y = 1, z = 1/2$. $\endgroup$
    – River Li
    Commented Mar 19, 2022 at 0:44

3 Answers 3


As an aside, reversing (or repeating) the change of variables further simplifies the work.

Starting off similar to OP's / River Li's work, we WTS

$$\sqrt{ ( x+y+z) (\frac{x}{x+y} + \frac{y}{y+z} + \frac{z}{z+x}) } \leq \frac{ xy+yz+zx} { \sqrt{2} }. $$

This is equivalent to

$$ \sum x + \sum \frac{xz}{x+y} \leq \frac{ (xy+yz+zx)^2 } { 2}.$$

We revert the change of variables, letting $ x = \frac{1}{a}$ and using $abc = 1$. We WTS

$$ \sum bc + \sum \frac{ ab^2}{a+b} \leq \frac{ (a+b+c)^2}{2} \Leftrightarrow \sum \frac{ab^2}{a+b} \leq \frac{ a^2+b^2+c^2}{2}. $$

This is true because

$$ \frac{ ab^2}{a+b} \leq \frac{ b^2+ab}{4} \Rightarrow \sum \frac{ab^2}{a+b} \leq \sum \frac{b^2+ab}{4} \leq \sum \frac{ a^2}{2}.$$


  • And of course, for those who don't want to substitute twice, you can work in just $a, b, c$. However, the steps seem "less obvious".
  • $\begingroup$ Yes, it is simpler. $\endgroup$
    – River Li
    Commented Mar 19, 2022 at 14:32

Remark: As Calvin Lin pointed out, we can just deal with $a, b, c$, without the substitutions.

We have \begin{align*} &\sum_{\mathrm{cyc}} \sqrt{\frac{ab}{bc^2 + 1}} \\ =\,& \sum_{\mathrm{cyc}} \sqrt{\frac{ab ab}{(bc^2 + 1)ab}}\\ =\,& \sum_{\mathrm{cyc}}\frac{ab}{\sqrt{ab + bc}}\\ \le\,& \sqrt{(ab + bc + ca)\left(\frac{ab}{ab + bc} + \frac{bc}{bc + ca} + \frac{ca}{ca + ab}\right)} \tag{1}\\[5pt] =\,&\sqrt{\frac{(ab + bc + ca)ab}{ab + bc} + \frac{(ab + bc + ca)bc}{bc + ca} + \frac{(ab + bc + ca)ca}{ca + ab}}\\[5pt] =\,& \sqrt{ab + \frac{ca^2}{a + c} + bc + \frac{ab^2}{b + a} + ca + \frac{bc^2}{c + b}}\\[5pt] \le\,& \sqrt{ab + \frac{\frac{(a + c)^2}{4}a}{a + c} + bc + \frac{\frac{(b + a)^2}{4}b}{b + a} + ca + \frac{\frac{(b + c)^2}{4}c}{c + b}}\tag{2}\\[5pt] =\,&\sqrt{\frac{1}{4}(a^2 + b^2 + c^2) + \frac54(ab + bc + ca)} \end{align*} where we have used the Cauchy-Bunyakovsky-Schwarz inequality in (1), and $ca \le \frac{(c + a)^2}{4}$ etc. in (2).

It suffices to prove that $$\frac{(a + b + c)^2}{2} \ge \frac{1}{4}(a^2 + b^2 + c^2) + \frac54(ab + bc + ca)$$ or $$a^2 + b^2 + c^2 \ge ab + bc + ca$$ which is true.

We are done.

  • $\begingroup$ How can you come up with the idea of transforming $\dfrac{zx}{x+y}$ into $\dfrac{zxy}{y(x+y)}$, I think this is the key of the problem. Are you aiming to make $(x+y)$ disappear in the denominator? $\endgroup$
    – trungbk
    Commented Mar 19, 2022 at 4:35
  • $\begingroup$ @trungbk Yes, we need to eliminate the denominator $(x+y)$. $\endgroup$
    – River Li
    Commented Mar 19, 2022 at 5:06
  • $\begingroup$ @trungbk Will you post $$a b \sqrt{a b}+b c \sqrt{b c}+c a \sqrt{c a} \leqslant a b c+\frac{1}{2} \sqrt[3]{\frac{\left(a^{2}+b c\right)^{2}\left(b^{2}+c a\right)^{2}\left(c^{2}+a b\right)^{2}}{a b c}}$$ in this site (with your attempt)? $\endgroup$
    – River Li
    Commented Mar 19, 2022 at 5:14
  • $\begingroup$ I wanted to post it, but the expression $\sum ab\sqrt{ab}$ and $(a^2+bc)(b^2+ca)(c^2+ab)$ is really strange to me, I don't have a working idea $\endgroup$
    – trungbk
    Commented Mar 19, 2022 at 11:42
  • $\begingroup$ Oooh, I didn't realize yours was essentially the same as mine. I was too lazy to wade through the algebra, and didn't like your (previous) final inequality lol. $\endgroup$
    – Calvin Lin
    Commented Mar 19, 2022 at 19:23

Use this inequality:

If a, b and c are all positive, show that:

$$\frac a{b+c}+\frac b{a+c}+\frac c{b+a}\geq\frac 32$$


Since a, b and c are positive we have a+b, b+c, and c+a are also positive. Let b+c<c+a<a+b, so that $\frac 1{b+c}\geq \frac 1{c+a} \geq \frac 1{a+b}$


$(a_1+a_2+a_3+...+a_n)(b_1+b_2+b_3+...b_n)\geq n(a_1b_1+a_2b_2+a_3b_3+ ...a_nb_n)$

we have:

$[(a+b)+(b+c)+(a+c)]\cdot \big[\frac 1{ a+b} +\frac 1{b+c}+\frac 1{a+c}\big]\geq3\big[ (a+b)\cdot \frac 1 {a+b}+(b+c)\cdot \frac 1 {b+c}+(a+c)\cdot \frac 1 {a+c}\big]$


$2(a+b+c)\big[\frac 1{a+b}+\frac 1{b+c} +\frac 1 {c+a}\big]\geq 9$


$\big(\frac a{b+c}+1+\frac b{c+a}+1+\frac c{a+b}+1\big)\geq \frac 9 2$


$\big(\frac a{b+c}+\frac b{c+a}+\frac c{a+b}\big)\geq \frac 3 2$

Now since you let $a=\frac 1x$, $b=\frac 1y$ and $c=\frac 1z$, the conditions changes and you have to get what you wanted to prove, i. e:

$\big(\frac x{x+y}+\frac y{y+z}+\frac z{z+x}\big)\leq \frac 3 2$

  • $\begingroup$ -1 As River Li noted, the final inequality isn't true. $\quad$ Can you elaborate on how your substitution works? I'm guessing that you made an error, esp since $ a/(b+c) -> yz/x(y+z) $ $\endgroup$
    – Calvin Lin
    Commented Mar 19, 2022 at 14:05

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