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Let $a$, $b$ and $c$ be non-negative numbers. Prove that: $$a\sqrt{a^2+bc}+b\sqrt{b^2+ac}+c\sqrt{c^2+ab}\geq\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}.$$

I have a proof, but my proof is very ugly:

it's enough to prove a polynomial inequality of degree $15$.

I am looking for an easy proof or maybe a long, but a smooth proof.

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  • 2
    $\begingroup$ If you assume $abc=1$, then $a^2+bc=a+a^{-1}$ $\endgroup$ – Empy2 Jan 1 '16 at 10:58
  • $\begingroup$ What is a rest? $\endgroup$ – Michael Rozenberg Jan 1 '16 at 10:59
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    $\begingroup$ No, it was just a thought, I don't have a full proof. $\endgroup$ – Empy2 Jan 1 '16 at 11:00
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    $\begingroup$ I tried smth, but getting a contradiction from it...: Let $abc=1$. $\sqrt{x^2+\frac1x}$ is convex (see W/A) so by weighted Jensen, $LHS\geq\sqrt{\frac{\sum a(a^2+\frac1a)}{\sum a}}\cdot\sum a=\sqrt{(a+b+c)(a^3+b^3+c^3+3abc)}$, but for $a\to0$ the reverse inequality is true (by e.g. Cauchy-Schwarz), so what did I do wrong? $\endgroup$ – rabota Jan 1 '16 at 12:25
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    $\begingroup$ To barto. It's wrong because $f(x)=\sqrt x$ is a concave function. $\endgroup$ – Michael Rozenberg Jan 1 '16 at 17:48
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This inequality is beautiful and tight. Equality holds when $a=b=c$, or $a=b$ and $c=0$.

Now we use the S.O.S. (sum of squares) technique to solve it (about this technique, see Sos - sum of squares for details).

First, squaring on both sides, it suffices to show

$$\sum a^2 \left( a^2+bc \right) + 2\sum bc\sqrt{\left( b^2+ca \right) \left( c^2+ab \right)} \ge 2\left( \sum a^2 \right) \left( \sum bc \right)\\ \Longleftrightarrow \sum a^4 + abc \sum a + 2\sum bc\sqrt{\left( b^2+ca \right) \left( c^2+ab \right)} \ge 2\sum bc\left( b^2+c^2 \right) + 2abc \sum a.$$

Note (from the formulation of Schur's inequality) that $$\sum \left( b^2 + c^2 - a^2 \right) (b-c)^2 = 2 \sum a^4 - 2 \sum bc\left( b^2 + c^2 \right) + 2abc\sum a.$$

Thus the inequality is equivalent to

$$\frac{1}{2} \sum \left( b^2 + c^2 - a^2 \right)(b-c)^2 + 2\sum bc \sqrt{\left( b^2+ca \right) \left( c^2+ab \right)} \ge \sum bc\left( b^2 + c^2 \right) + 2abc \sum a.$$

Note that

$$\sum bc \left( \frac{(b-c)(b+c-a)}{\sqrt{b^2+ca} + \sqrt{c^2+ab}} \right)^2 = \sum bc \left( \sqrt{b^2+ca} - \sqrt{c^2+ab} \right)^2\\ = \sum bc \left( \left( b^2+ca \right) + \left( c^2+ab \right) \right) - 2 \sum bc \sqrt{\left( b^2+ca \right) \left( c^2+ab \right)}\\ = \sum bc \left( b^2+c^2 \right) + 2abc \sum a - 2 \sum bc \sqrt{\left( b^2+ca \right) \left( c^2+ab \right)}.$$

Thus the inequality is equivalent to

$$\sum \left( b^2+c^2-a^2 - \frac{2bc(b+c-a)^2}{\left( \sqrt{b^2+ca} + \sqrt{c^2+ab} \right)^2} \right) (b-c)^2 \ge 0.$$

Note that

$$\left( \sqrt{b^2 + ca} + \sqrt{c^2 + ab} \right)^2 = b^2+ca+c^2+ab + 2\sqrt{\left( b^2+ca \right) \left( c^2+ab \right)}\\ \ge b^2+ca+c^2+ab+2bc=(b+c)(a+b+c),$$

and similarly

$$\left( \sqrt{c^2 + ab} + \sqrt{a^2 + bc} \right)^2 \ge (c+a)(a+b+c)\\ \left( \sqrt{a^2 + bc} + \sqrt{b^2 + ca} \right)^2 \ge (a+b)(a+b+c).$$

So if we define

$$S_a = b^2+c^2-a^2 - \frac{2bc(b+c-a)^2}{(b+c)(a+b+c)},\\ S_b = c^2+a^2-b^2 - \frac{2ca(c+a-b)^2}{(c+a)(a+b+c)},\\ S_c = a^2+b^2-c^2 - \frac{2ab(a+b-c)^2}{(a+b)(a+b+c)},$$

then we only need to show $\sum S_a (b-c)^2 \ge 0$. Without loss of generality, we assume that $a \ge b \ge c$. In a post by hungkhtn in Sos - sum of squares, we know that it suffices to show

$$S_b \ge 0,\ S_c\ge 0,\ a^2 S_b + b^2 S_a \ge 0.$$

First,

$$\left( c^2+a^2-b^2 \right)-(c+a-b)^2=2b(c+a)-2ca-2b^2=-2(b-a)(b-c)\ge 0\\ \Longrightarrow \left( c^2+a^2-b^2 \right) (c+a)(a+b+c)\ge (c+a-b)^2\cdot (c+a)(a+b+c)\ge 2ca(c+a-b)^2,$$

which means $S_b\ge 0$.

Besides,

$$\left( a^2+b^2-c^2 \right)(a+b)(a+b+c) - 2ab(a+b-c)^2\\ = \left(a^2+b^2\right) (a+b)^2 + \left(a^2+b^2\right)(a+b)c- (a+b)^2 c^2 - (a+b)c^3\\ - 2ab(a+b)^2 + 4abc(a+b) - 2abc^2\\ = (a-b)^2 (a+b)^2 + \left( \left(a^2+b^2\right)(a+b)c - (a+b)^2 c^2 \right) + \left( 4abc(a+b) - 2abc^2 - (a+b)c^3 \right)\\ \ge 0,$$

which means $S_c \ge 0$.

Finally,

$$a^2 S_b + b^2 S_a \ge 0\\ \Longleftrightarrow a^2 c^2 + a^4 - a^2 b^2 + b^4 + b^2 c^2 - b^2 a^2 \ge \frac{2ca^3(c+a-b)^2}{(c+a)(a+b+c)} + \frac{2b^3 c(b+c-a)^2}{(b+c)(a+b+c)}\\ \Longleftrightarrow (a+b+c)\left( (a+b)^2 (a-b)^2 + \left( a^2+b^2 \right) c^2 \right) \ge \frac{2ca^3 (c+a-b)^2}{c+a} + \frac{2cb^3 (c+b-a)^2}{c+b}.$$

Note that the right hand side is not greater than

$$2ca^2 (c+a-b)^2 + 2cb^2 (c+b-a)^2\\ = \left( 2c^3 a^2 + 4c^2 a^2 (a-b) + 2ca^2 (a-b)^2 \right) + \left( 2c^3 b^2 + 4c^2 b^2 (b-a) + 2cb^2 (a-b)^2 \right)\\ = 2c^3 \left( a^2 + b^2 \right) + 4c^2 (a+b)(a-b)^2 + 2c\left( a^2 + b^2 \right)(a-b)^2\\ \le (a+b+c)\left(a^2+b^2\right)c^2 + 4c^2 (a+b)(a-b)^2 + 2c\left( a^2 + b^2 \right)(a-b)^2.$$

So it remains to show

$$(a+b+c)(a+b)^2 \ge 4c^2(a+b) + 2c\left( a^2+b^2 \right)\\ \Longleftrightarrow (a+b)^3 \ge 4c^2 (a+b) + c(a-b)^2\\ \Longleftrightarrow (a+b)\left( (a+b)^2 - 4c^2 \right) \ge c(a-b)^2,$$

which is obviously true, since $(a+b)^2 - 4c^2 \ge (a+b)^2 - 4ab = (a-b)^2$ and $a+b\ge c$.

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$\sum\limits_{cyc}a\sqrt{a^2+bc}\geq\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}\Leftrightarrow$

$\Leftrightarrow\sum\limits_{cyc}\left(a^4+a^2bc+2ab\sqrt{(a^2+bc)(b^2+ac)}\right)\geq\sum\limits_{cyc}(2a^3b+2a^3c+2a^2bc)\Leftrightarrow$

$\sum\limits_{cyc}(a^4-a^3b-a^3c+a^2bc)\geq\sum\limits_{cyc}\left(a^3b+a^3c+2a^2bc-2ab\sqrt{(a^2+bc)(b^2+ac)}\right)\Leftrightarrow$

$\Leftrightarrow\frac{1}{2}\sum\limits_{cyc}(a-b)^2(a+b-c)^2\geq\sum\limits_{cyc}ab\left(a^2+bc+b^2+ac-2\sqrt{(a^2+bc)(b^2+ac)}\right)\Leftrightarrow$

$\Leftrightarrow\sum\limits_{cyc}(a-b)^2(a+b-c)^2\geq2\sum\limits_{cyc}ab\left(\sqrt{a^2+bc}-\sqrt{b^2+ac}\right)^2\Leftrightarrow$

$\Leftrightarrow\sum\limits_{cyc}(a-b)^2(a+b-c)^2\left(1-\frac{2ab}{\left(\sqrt{a^2+bc}+\sqrt{b^2+ac}\right)^2}\right)\geq0$, which is obvious.

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protected by Community Feb 12 '16 at 17:31

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