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Let $a,b,c$ be the side-lengths of a triangle. Prove that:

I. $$\sum_{cyc}{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$$

What I have tried:

\begin{eqnarray} a-b+c&=&x\\ b-c+a&=&y\\ c-a+b&=&z. \end{eqnarray} So $a+b+c=x+y+z$ and $2a=x+y$, $2b=y+z$, $2c=x+z$ and our inequality become:

$$\sum_{cyc}{\frac{\sqrt{x}\cdot(x+2y+z)\cdot(x+y+2)}{4}} \geq 4\cdot(x+y+z)\cdot\sqrt{xyz}. $$ Or if we make one more notation : $S=x+y+z$ we obtain that:

$$\sum_{cyc}{\sqrt{x}(S+y)(S+z)} \geq 16S\cdot \sqrt{xyz} \Leftrightarrow$$

$$S^2(\sqrt{x}+\sqrt{y}+\sqrt{z})+S(y\sqrt{x}+z\sqrt{x}+x\sqrt{y}+z\sqrt{y}+x\sqrt{z}+y\sqrt{z})+xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y} \geq 16S\sqrt{xyz}.$$ To complete the proof we have to prove that:

$$y\sqrt{x}+z\sqrt{x}+x\sqrt{y}+z\sqrt{y}+x\sqrt{z}+y\sqrt{z} \geq 16\sqrt{xyz}. $$

Is this last inequality true ?

II.

Knowing that: $$p=\frac{a+b+c}{2}$$ we can rewrite the inequality: $$\sum_{cyc}{(2p-c)(2p-a)\sqrt{2(p-b)}} \geq 8p \sqrt{2^3 \cdot (p-a)(p-b)(p-c)} \Leftrightarrow$$

$$\sum_{cyc}{(2p-c)(2p-a)\sqrt{(p-b)}} \geq 16p \sqrt{(p-a)(p-b)(p-c)}$$

This help me ?

Thanks :)

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Re the proposed inequality $y \sqrt{x} + \cdots + y \sqrt{z} \geq 16 \sqrt{xyz}$: This is false when $x=y=z$. You have an algebra error going from $S^2 (\sqrt{x}+\sqrt{y}+\sqrt{z}) + \cdots \geq 16 S \sqrt{xyz}$ to $y \sqrt{x} + z \sqrt{x} + \cdots \geq 16 \sqrt{xyz}$. –  David Speyer Sep 13 '12 at 13:49

2 Answers 2

up vote 3 down vote accepted

Notice that the inequality proposed is proved once we estabilish

$$\frac{(a+b)(b+c)}{\sqrt{a+b-c}\sqrt{-a+b+c}}+\frac{(b+c)(c+a)}{\sqrt{a-b+c}\sqrt{-a+b+c}}+\frac{(c+a)(a+b)}{\sqrt{a-b+c}\sqrt{a+b-c}}\geq 4(a+b+c).$$

Using AM-GM on the denominators in the LHS, we estabilish that

  • $\sqrt{a+b-c}\sqrt{-a+b+c}\leq b$,
  • $\sqrt{a-b+c}\sqrt{-a+b+c}\leq c$,
  • $\sqrt{a-b+c}\sqrt{a+b-c}\leq a$.

Therefore

$$\operatorname{LHS}\geq \frac{(a+b)(b+c)}{b}+\frac{(b+c)(c+a)}{c}+\frac{(c+a)(a+b)}{a}=3(a+b+c)+\frac{ca}{b}+\frac{ab}{c}+\frac{bc}{a}$$

To finish with the proof it suffices then to prove

$$\frac{ca}{b}+\frac{ab}{c}+\frac{bc}{a}\geq a+b+c.$$

This follows from AM-GM, indeed

  • $$\frac{\frac{ac}{b}+\frac{ab}{c}}{2}\geq a,$$
  • $$\frac{\frac{bc}{a}+\frac{ab}{c}}{2}\geq b,$$
  • $$\frac{\frac{ac}{b}+\frac{bc}{a}}{2}\geq c;$$

summing up these last inequalities, we get the desired result, hence finishing the proof. Notice that the condition of $a,b,c$ being the sides of a triangle is essential in the first usage of AM-GM.

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@Iuli.. did you check this answer? Does it seem correct to you? –  uforoboa Sep 25 '12 at 19:29

I couldn't solve this with elementary means. Take $\sum_{cyc}{\sqrt{x}(S+y)(S+z)} \geq 16S\cdot \sqrt{xyz} $ and note that it is enough to prove it for $S=1$ because it is homogeneous. Substituting $x\to x^2$, etc. consider the function $f(x,y,z)=(x^2+1) (y^2+1) z+(x^2+1) y (z^2+1)+x (y^2+1) (z^2+1)-16 x y z$. It is enough to prove it is non-negative if $\sum_{cyc}{x^2}=1$.

Let's denote $p[i]=x^i+y^i+z^i$.

Now notice that the following decomposition is valid: $$12f(x,y,z)=12 p[1] - 32 p[1]^3 + p[1]^5 + 108 p[1] p[2] - 4 p[1]^3 p[2] + 3 p[1] p[2]^2 - 76 p[3] + 2 p[1]^2 p[3] - 2 p[2] p[3]=$$

Using our assumption $p[2]=1$ this simplifies to:$f(x,y,z)=123 p[1] - 36 p[1]^3 + p[1]^5 - 78 p[3] + 2 p[1]^2 p[3]$.

Lets put $x=p[1]$ and $a=p[3]$.

Because $p[2]=1$ we ca deduce the ranges for $a$ and $x$ as $1 \ge p[3] \ge 1/\sqrt{3}$ and $ \sqrt{3}\ge p[1] \ge 1$. Also we need to keep in mind that from the power means inequality $p[1]/3 \le \sqrt[3]{p[3]/3}$.

We need to prove that the function:$g(x)=123 x - 36 x^3 + x^5 - 78 a + 2 a x^2 $ is non-negative in our domain. It turns out that this problem is numerically tractable, and this function is decreasing within our domain and becomes zero only when $a=\sqrt{3}/3$, and $x=\sqrt{3}$. The calculations however are quite tedious.

I am sure that a better way to solve this exists.

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