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

Question

If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$. Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.

Answer 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, $$\triangle= -4\cdot1\cdot\left(-\frac12\right)^3-27(-u)^2=\frac12-27u^2\ge0\iff u^2\le \frac1{54}$$

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

Answer 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?

Answer 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 $$ (abc)^2\le \frac{K^6}{16}=\frac{8/27}{16}=\frac1{54}. $$

Answer 5

A Lagrange multiplier proof:

Suppose $a^2b^2c^2$ is maximized. Then since $a + b + c = 0$, two of $a,b$, and $c$ are of one sign, and the third is of the other sign. (Clearly none are zero). Replacing $a,b,$ and $c$ by their negatives if needed, one can assume two are positive, and permuting the variables if necessary one can assume $a,b > 0$ and $c < 0$. Then since $c = -(a + b)$, the equation $a^2 + b^2 + c^2 = 1$ becomes $a^2 + b^2 + (a + b)^2 = 1$ or $$a^2 + ab + b^2 = {1 \over 2}$$ And one wants to maximize $a^2b^2c^2$ subject to this contraint for $a, b > 0$. Since $c = -(a + b) < 0$ this is equivalent to maximizing $ab(a + b)$ subject to the constraint, for $a, b > 0$. The Lagrange conditions are $$2ab + b^2 = \lambda(2a + b)$$ $$a^2 + 2ab = \lambda(a + 2b)$$ Since $2ab + b^2 = b(2a + b)$, the first equation says $\lambda = b$. Similarly, the second equation implies $\lambda = a$. So $a = b$, and then $a^2 + ab + b^2 = {1 \over 2}$ means $3a^2 = {1 \over 2}$ or $a = {1 \over \sqrt{6}}$. So $a = b = {1 \over \sqrt{6}}$. Then $c = -(a + b) = -{2 \over \sqrt{6}}$, and $a^2b^2c^2 = {4 \over 6^3} = {1 \over 54}$.

Answer 6

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

$$\implies 0^2=1+2 (ab+bc+ca)\implies ab+bc+ca=-\frac12$$

Again, $ab+bc+ca=a(b+c)+bc=a(-a)+bc\implies bc-a^2$

$$\implies bc-a^2=-\frac12\implies bc=\frac{2a^2-1}2$$

$$\implies abc=a\cdot \frac{2a^2-1}2=\frac{2a^3-a}2$$

Now use Second derivative test for the extreme values of $f(a)=\frac{2a^3-a}2$

As $a^2=1-(b^2+c^2)\le1, -1\le a\le 1$

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Alternatively, we can set $a=\cos A, b=\sin A\cos B,c=\sin A\sin B$

As $a+b+c=0,\cos A+\sin A\cos B+\sin A\sin B=0\implies \cos A=-\sin A(\cos B+\sin B) $

Squaring we get, $\cot^2A=1+2\sin B\cos B$

$$abc=\cos A\cdot\sin A\cos B\cdot\sin A\sin B=\cos A\sin^2A(\sin B\cos B)$$ $$=\cos A\sin^2A\cdot \frac{\cot^2A-1}2=\cos A\cdot\frac{\cos^2A-\sin^2A}2=\frac{2\cos^3A-\cos A}2 $$ which can be handled like the previous method

Answer 7

Form an equation with $a,b,c$ as its roots.

We have $ab+bc+ca=-\frac{1}{2}$ from the given relation.

So the equation with $a,b,c$ as its roots is,

$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0$

$\Rightarrow x^3-\frac{1}{2}x-abc=0$

If all of $a,b,c$ are of same sign then we have $a=b=c=0$ in which case the inequality is trivially true.So we may consider without loss of generality that $c>0,b<0$ then we have

$a=-(c+b)\Rightarrow a^2=c^2+b^2+2bc\le c^2+b^2=1-a^2\Rightarrow 2a^2\le 1\Rightarrow \frac{1}{2}-a^2\ge 0$

Putting $a$ in the above equation we have $a^3-\frac{1}{2}a+abc=0$

So we have ,

$abc=a(a^2-\frac{1}{2})$

$\Rightarrow a^2b^2c^2=a^2(a^2-\frac{1}{2})^2=a^2(\frac{1}{2}-a^2)^2$

By applying A.M. G.M. on $2a^2(\frac{1}{2}-a^2)^2$(Note that $a^2\ge 0,\frac{1}{2}-a^2\ge 0$ so we can apply A.M.-G.M.) we have,

$2a^2(\frac{1}{2}-a^2)^2\le \left(\frac{2a^2+\frac{1}{2}-a^2+\frac{1}{2}-a^2}{3}\right)^{3}=\frac{1}{27}\Rightarrow a^2b^2c^2=a^2(\frac{1}{2}-a^2)^2\le \frac{1}{27}$

Equality will hold iff $2a^2=\frac{1}{2}-a^2\Rightarrow 3a^2=\frac{1}{2}\Rightarrow a=\frac{1}{\sqrt{6}}$

Answer 8

As several have shown above, a, b and c are roots of the cubic $abc=t^3-.5t$. In order for this to have three real roots, abc must lie between the maximum and minimum of $t^3-.5t$. At one of these extrema, two of the roots are equal to the location of the extrema, and the third is found from the requirement that the sum of the roots be zero. One set is $a=b=\frac{1}{\sqrt{6}}$, $c=\frac{-2}{\sqrt{6}}$. Notice that it is not necessary to calculate the discriminant, which can sometimes be messy if the sum of the roots is nonzero.