If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger inequality. Thanks.

  • $\begingroup$ you are missing a $c$ in the last term. $\endgroup$ – Chinny84 Aug 7 '15 at 16:23
  • $\begingroup$ How about starting with $a=8\tan^2A$ etc. where $0<A<\dfrac\pi2$ $\endgroup$ – lab bhattacharjee Aug 7 '15 at 16:42
  • $\begingroup$ I don't even understand the notation. Care to explain it? $\endgroup$ – Harald Hanche-Olsen Jul 18 '17 at 10:52
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    $\begingroup$ @HaraldHanche-Olsen $\sum_{cyc}$ is a pretty common piece of notation in recreational inequalities. It means "Over all cyclic permutations of the ordered triple $(a, b, c)$" (you're expected to understand the variable names from context, and while it's not restricted to three variables, inequalities usually have three variables). For instance, $\sum_{cyc}a$ is the same as $a+b+c$, while $\sum_{cyc}\frac ab$ is the same as $\frac ab + \frac bc + \frac ca$. $\endgroup$ – Arthur Jul 18 '17 at 10:55
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    $\begingroup$ This problem is very similar to one of IMO problems. $\endgroup$ – Michael Rozenberg Jul 18 '17 at 11:03

First let $$ x = \sqrt{\frac{a}{a+8}}, \,\, y = \sqrt{\frac{b}{b+8}}, \,\, z = \sqrt{\frac{c}{c+8}} \,\, $$ Then $1 > x,y,z > 0$ and $$ a = \frac{8x^2}{1 - x^2}, \,\, b = \frac{8y^2}{1 - y^2}, \,\, c = \frac{8z^2}{1 - z^2},\,\, $$

So the question transforms to this:

Given that $1 > x,y,z > 0, \, \, \frac{512x^2y^2z^2}{(1 - x^2)(1 - y^2)(1 - z^2)} = 1$, prove that $x + y + z \geqslant 1$.

Prove this by contradiction. Suppose on the contrary that $x + y + z < 1$, then

$$ \begin{align} (1 - x^2)(1 - y^2)(1 - z^2) &= (1 - x)(1 + x)(1 - y)(1 + y)(1 - z)(1 + z) \\ &>(x + x + y + z)(y + z)(x + y + y + z)(x + z)(z + x + y + z)(x + y) \\ &\geqslant 4x^{\frac12}y^{\frac14}z^{\frac14}\cdot 2y^{\frac12}z^{\frac12} \cdot 4y^{\frac12}x^{\frac14}z^{\frac14}\cdot 2x^{\frac12}z^{\frac12} \cdot 4z^{\frac12}y^{\frac14}x^{\frac14}\cdot 2y^{\frac12}x^{\frac12}\\ &=512 x^{\frac12 + \frac14 + \frac12 + \frac14 + \frac12}y^{\frac14 + \frac12 + \frac12 + \frac14 + \frac12}z^{\frac14 + \frac12 +\frac14 + \frac12 + \frac12} \\ &= 512x^2y^2z^2 \end{align}$$ And this is contradictory to the condition.


The required inequality is trivialized by the claim below. The equality case is when $a=b=c=1$.

Claim: If $a,b,c>0$ are such that $abc=1$, then $\displaystyle\sqrt{\frac{a}{a+8}}\geq \frac{a^{4/9}}{a^{4/9}+b^{4/9}+c^{4/9}}$. The equality holds if and only if $a=b=c=1$.

Proof: Note that the required inequality is equivalent to $$\left(a^{4/9}+b^{4/9}+c^{4/9}\right)^2 \geq a^{-1/9}(a+8)\,,$$ which is also equivalent to $$\left(b^{4/9}+c^{4/9}\right)\left(a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\right) \geq 8a^{-1/9}\,.$$ To prove the previous inequality, we invoke the AM-GM Inequality twice: $$b^{4/9}+c^{4/9}\geq 2b^{2/9}c^{2/9}$$ and $$a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\geq 4a^{1/9}a^{1/9}b^{1/9}c^{1/9}=4a^{2/9}b^{1/9}c^{1/9}\,.$$ Thus, $$ \begin{align} \left(b^{4/9}+c^{4/9}\right)\left(a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\right) &\geq \left(2b^{2/9}c^{2/9}\right)\left(4a^{2/9}b^{1/9}c^{1/9}\right) \\ &=8a^{2/9}b^{1/3}c^{1/3}=8a^{-1/9}\left(abc\right)^{1/3}=8a^{-1/9}\,, \end{align}$$ which is what we want. By the equality condition of the AM-GM Inequality, the equality happens iff $a=b=c=1$.

P.S.: I just realized why this inequality looks so familiar. It is equivalent to IMO'2001#2 (http://imo.wolfram.com/problemset/IMO2001_solution2.html). Substitute $a$, $b$, and $c$ by $x^3$, $y^3$, and $z^3$, then homogenize the required inequality via the condition $xyz=1$, and you will see what I'm talking about.


Let $a=\frac{x^2}{yz}$, $b=\frac{y^2}{xz}$ and $c=\frac{z^2}{xy}$, where $x$, $y$ and $z$ are positives.

Hence, by Holder and AM-GM we obtain: $$\sum_{cyc}\sqrt{\frac{a}{a+8}}=\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x}{\sqrt{x^2+8yz}}\right)^2\sum\limits_{cyc}x(x^2+8yz)}{\sum\limits_{cyc}x(x^2+8yz)}}\geq$$ $$\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+8zyz)}}\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+3x^2y+3x^2z+2xyz)}}=1.$$ Done!


Let $a=\frac{x^2}{yz}$ and $b=\frac{y^2}{xz}$, where $x$, $y$ and $z$ are positives.

Hence, $c=\frac{z^2}{xy}$ and we need to prove that: $$\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}\geq1.$$ Now, by Holder $$\left(\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}\right)^2\sum_{cyc}x(x^2+8yz)\geq(x+y+z)^3.$$ Thus, it remains to prove that $$(x+y+z)^3\geq\sum_{cyc}x(x^2+8yz)$$ or $$\sum_{cyc}z(x-y)^2\geq0.$$ Done!

Another way: $$\sum_{cyc}\sqrt{\frac{a}{a+8}}=\sum_{cyc}\frac{x}{\sqrt{x^2+8xy}}\geq\sum_{cyc}\frac{x^{\frac{4}{3}}}{x^{\frac{4}{3}}+y^{\frac{4}{3}}+z^{\frac{4}{3}}}=1$$

Also we can use the Contradiction method.

Let $\frac{a}{a+8}=\frac{p^2}{9}$, $\frac{b}{b+8}=\frac{q^2}{9}$ and $\frac{c}{c+8}=\frac{q^2}{9}$, where $p$, $q$ and $r$ are positives.

Hence, we need to prove that $p+q+r\geq3.$

But the condition $abc=1$, gives $$8^3=\prod_{cyc}\left(\frac{9}{p^2}-1\right)$$ or $$81=57p^2q^2r^2-p^2q^2-p^2r^2-q^2r^2+9(p^2+q^2+r^2).$$ Now, let $p+q+r<3$, $p=kx$, $q=y$ and $r=z$, where $k>0$ and $x+y+z=3$.

Hence, $kx+y+z<3=x+y+z$, which gives $0<k<1$.

Thus, since $9-p^2-q^2>(p+q)^2-p^2-q^2>0$, we obtain: $$81=57p^2q^2r^2-p^2q^2-p^2r^2-q^2r^2+9(p^2+q^2+r^2)=$$ $$=k^2x^2(57y^2z^2-y^2-z^2+9)-y^2z^2+9(y^2+z^2)<$$ $$<x^2(57y^2z^2-y^2-z^2+9)-y^2z^2+9(y^2+z^2),$$ which is contradiction because we'll prove now that $$57x^2y^2z^2-x^2y^2-x^2z^2-y^2z^2+9(x^2+y^2+z^2)\leq81.$$ Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Thus, it's obvious that our inequality is equivalent to $f(w^3)\leq0$, where $f$ is a convex function.

Id est, it's enough to prove the last inequality for an extremal value of $w^3$,

which happens in the following cases.

  1. $y=x$, $z=3-2x$, where $0<x<1.5$, which gives $$x(x-1)^2(9+15x+19x^2-19x^3)\geq0,$$ which is true for $0<x<\frac{3}{2}$;

  2. $w^3=0$.

Let $z=0$ and $y=3-x$, where $0<x<3$.

We obtain, $x(4x^3-12x^2-9x+54)\geq0$, which is obvious.

Done again!

  • $\begingroup$ Thank you. How can we know what we should substitute the variables with, trial and error ? $\endgroup$ – carat Jul 18 '17 at 12:03
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    $\begingroup$ @carat Yes, sometimes it's with trial and error, but in our case there are many substitutions. Two of them you saw. There is also $a=e^x$, $b=e^y$ and $c=e^z$, which gives a fourth proof. $\endgroup$ – Michael Rozenberg Jul 18 '17 at 12:30
  • $\begingroup$ I've learned a lot. Thank you so much for your kind help. $\endgroup$ – carat Jul 18 '17 at 12:50
  • $\begingroup$ @carat You are welcome! $\endgroup$ – Michael Rozenberg Jul 18 '17 at 12:52

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