$$\frac{a+b}{c} + \frac{b+c}{a}+\frac{a+c}{b} + 6 \ge 2\sqrt2(\sqrt\frac{1-a}{a} + \sqrt\frac{1-b}{b} + \sqrt\frac{1-c}{c})$$

The condition set for this is : $a + b + c = 1$

From that condition I tried to rewrite as : $$\frac{a+b}{c} + \frac{b+c}{a}+\frac{a+c}{b} + 6a + 6b + 6c \ge 2\sqrt2(\sqrt\frac{1-a}{a} + \sqrt\frac{1-b}{b} + \sqrt\frac{1-c}{c})$$

Then I tried to perform $AM-GM$ tricks which I failed to find anything useful.

I tried things such as: $$\frac{a+b}{c} + 6c \ge 2\sqrt{6(a+b)}$$ or $$\frac{a}{c} + 3c \ge 2\sqrt{3a}$$ which if summed side by side would equal : $$LHS \ge 4\sqrt{3a} + 4\sqrt{3b} + 4\sqrt{3c}$$

From what I have to prove , it seems that I need to get on the RHS a big square root of the factor 8 and factor $(\sqrt\frac{1-a}{a} + \sqrt\frac{1-b}{b} + \sqrt\frac{1-c}{c})$ but I don't see any methotds how.

Any kind of help would be great and thank you in advance.

  • 1
    $\begingroup$ @coffeemath Note the problem implicitly requires $a$, $b$ and $c$ to be between $0$ and $1$ for $\frac{1-x}{x}$, with $x$ being $a$, $b$ and $c$, to all be positive and, thus, for each to have a real square root. Nonetheless, this requirement should be explicitly stated. $\endgroup$ Sep 20 at 19:33

1 Answer 1


From a+b+c=1 we get that a+b=1-c etc, so the LHS is equal to $\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}+6$. We can use the notation $x=\frac{1-a}{a}, y=\frac{1-b}{b}, z=\frac{1-c}{c}$ and now the inequality can be written as $x+y+z+6 \geq 2\sqrt{2}(\sqrt{x}+\sqrt{y}+\sqrt{z})$. Multiplying the RHS and moving everything to the LHS, we get $(x-2\sqrt{2x}+2) +(y-2\sqrt{2y}+2) +(z-2\sqrt{2z}+2) \geq 0$, which is equivalent to $(\sqrt{x}-\sqrt{2})^2+(\sqrt{y}-\sqrt{2})^2+(\sqrt{z}+\sqrt{2})^2 \geq 0$, which is true. Given the fac that $a+b+c=1$ and $\sqrt{\frac{1-a}{a}}$ is defined, we don't have to check the boundries of $x,y,z$.


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