$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ [closed]

Solution 1:

So, $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$

$\implies ab+bc+ca=-\frac12.$

So $a,b,c$ are the roots of the equation $t^3-\frac12t-u=0$ where $u=abc.$

Using discriminant of the cubic equation fro real roots, $$\triangle= -4\cdot1\cdot\left(-\frac12\right)^3-27(-u)^2=\frac12-27u^2\ge0\iff u^2\le \frac1{54}$$

Solution 2:

$c=-a-b$, so $$a^2+b^2+c^2=2a^2+2ab+2b^2=1$$ let $a=x+y,b=x-y$ then $$3x^2+y^2=1/2$$ with the we only consider $ab\le 0$ and $c=-(a+b)=-2x$, $$c^2=a^2+2ab+b^2\le a^2+b^2=1-c^2$$ so $$8x^2\le 1$$ so $$abc=ab(-a-b)=-(a+b)ab=-2x(x^2-y^2)=-x(8x^2-1)$$

then $$a^2b^2c^2=x^2(8x^2-1)(8x^2-1)=\dfrac{1}{16}\cdot 16x^2(1-8x^2)(1-8x^2)\le \dfrac{1}{16}\dfrac{8}{27}=\dfrac{1}{54}$$

Solution 3:

There are three real numbers, and only two sign(negative, non-negative),So there are two real numbers with same sign,then without loss generality, Assume ab $\ge$ 0

$1 = a^2+b^2+c^2 = a^2+b^2+(-a-b)^2 = (a^2+b^2+ab) * 2 ≥ 3*ab * 2$
So $ab ≤ \dfrac{1}{6}$

$a^2 b^2 c^2 = a^2b^2(a^2+b^2+2ab)=a^2b^2(\dfrac{1}{2}+ab)≤\left(\dfrac{1}{6} \right)^2 *(\dfrac{1}{2}+\dfrac{1}{6})=\dfrac{1}{54}$

Hi, does someone give me any tips about expression format?

Solution 4:

Several good answers have been posted. Just feeling like adding a single variable solution with a geometric twist.

The set of points $(a,b,c)$ is the intersection of the unit sphere and the plane $T:x+y+z=0$ with normal vector $(1,1,1)$. IOW a circle of radius one in the plane $T$ centered at the origin. An occasionally useful factoid is that those points are parametrized by the formula $$ (a,b,c)=K(\cos t, \cos(t+\frac{2\pi}3), \cos(t-\frac{2\pi}3)), $$ with the constant $K=\sqrt{2/3}$. Some of you may have seen the relevant trig identities in high school physics in connection with three-phase power: the total power is constant (=> on the unit sphere) and the instantaneous sum of the voltages is zero (=> on the plane $T$).

Anyway, here $$ bc=K^2\cos(t+\frac{2\pi}3)\cos(t-\frac{2\pi}3)=K^2(\cos^2t-\frac34) $$ by basic trig identities. Therefore $$ abc=K^3\frac{4\cos^3t-3\cos t}4=\frac{K^3\cos3t}4. $$ Hence $$ (abc)^2\le \frac{K^6}{16}=\frac{8/27}{16}=\frac1{54}. $$