3

let $a,b,c\ge 0$, and such $$a+b+c=6$$

show that $$100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$$

My idea: since $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=36-2(ab+bc+ac)$$ $$a^2b^2+b^2c^2+a^2c^2=(ab+ca+bc)^2-2abc(a+b+c)=(ab+bc+ac)^2-12abc$$ so let $p=a+b+c=6, q=ab+bc+ac,r=abc$,so $$\Longleftrightarrow 100+5[36-2q]-2[q^2-12r]-r^2\ge 0$$ $$\Longleftrightarrow r^2+2q^2-24r+10q\le 280$$ then I can't.Thank you

Community
  • 1
math110
  • 93,304

2 Answers2

2

Let $G(a,b,c)=100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2$. We will show that $G(a,b,c) \geq G(2,2,2)=0$.

From the identity

$$ 4\frac{G(a,b,c)-G(a,\frac{b+c}{2},\frac{b+c}{2})}{(c-b)^2} +\frac{G(a,b,c)-G(a,b+c,0)}{bc}=\frac{(a^2+2)(b+c)^2}{4} $$

we see that at least one of the two numbers $G(a,b,c)-G(a,\frac{b+c}{2},\frac{b+c}{2})$ or $G(a,b,c)-G(a,b+c,0)$ is nonnegative, so that it suffices to treat the cases when one variable is zero or when two variables are equal. This is what we do in the next two lemmas.

Lemma 1. $G(a,b+c,0) \geq G(3,3,0)=28$.

Proof of lemma 1 Putting $d=b+c=6-a$, and $x=a-3$ we have

$$ \begin{array}{lcl} G(a,d,0) &=& 100+5(a^2+d^2)-2a^2d^2 \\ &=& 100+5((3+x)^2+(3-x)^2)-2((9-x^2)^2) \\ &=& 28+46x^2-2x^4 \geq 28=G(3,3,0). \end{array} $$

Lemma 2. $G(a,\frac{b+c}{2},\frac{b+c}{2}) \geq G(2,2,2)=0$.

Proof of lemma 2 Putting $e=\frac{b+c}{2}=\frac{6-a}{2}$, we have

$$ \begin{array}{lcl} G(a,e,e) &=& 100+5(a^2+2e^2)-2(2a^2e^2+e^4)-a^2e^4 \\ &=& 100+5\bigg(a^2+2\big(\frac{6-a}{2}\big)^2\bigg) -2\bigg(2a^2\bigg(\frac{6-a}{2}\bigg)^2+\bigg(\frac{6-a}{2}\bigg)^4\bigg) -a^2\bigg(\frac{6-a}{2}\bigg)^4 \\ &=& (a-2)^2\Bigg(7+a\bigg(\frac{7}{4}+\frac{9(6-a)}{8}+\frac{(6-a)^2}{8}+\frac{(6-a)^3}{16}\bigg) \bigg) \Bigg) \geq 0. \end{array} $$

Ewan Delanoy
  • 61,600
0

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, the condition does not depend on $w^3$ and we need to prove that $f(w^3)\geq0$,

where $f$ is a concave function.

But the concave function gets a minimal value for an extremal value of $w^3$,

which happens for an equality case of two variables or when maybe $w^3=0$.

Thus, it's enough to prove our inequality in the following cases.

  1. $b=a$ and $c=6-2a$, where $0\leq a\leq 3$, which gives $$(a-2)^2(a^2+5)(7+4a-2a^2)\geq0,$$ which is obvious for $0\leq a\leq 3$;

  2. $c=0$ and $b=6-a$, where $0\leq a\leq6$.

In this case we get $$140-30a-31a^2+12a^3-a^4\geq0,$$ which is true for $0\leq a\leq6$.

Done!

Michael Rozenberg
  • 194,933