Apply Vieta's?

First we are going to find $ab+ac+bc$ with the identity $(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$ :

$ab+ac+bc=-\dfrac{1}{2}$

Then consider a monic polynomial $P(x)$ with roots $a,b$ and $c$ . Then:

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

Substituting the known values:

$P(x)=x^3-\dfrac{1}{2}x-abc$

Now, for $a,b$ and $c$ to be real the discriminant of $P$ must be greater than or equal to zero. Also, the discriminant of $P(x)=x^3+px+q$ is just $\Delta=-4p^3-27q^2$ , so:

$\Delta=-4\left(-\dfrac{1}{2}\right)^3-27(-abc)^2 \geq 0$

$4\left(-\dfrac{1}{8}\right)+27(abc)^2 \leq 0$

$27(abc)^2 \leq \dfrac{1}{2}$

$(9abc)^2 \leq \boxed{\dfrac{3}{2}}$

$\text{Consider a monic cubic polynomial with roots as a,b,c}$

$P(x)=x^{3}+S_{1}x^{2}+S_{2}x+S_{3}$

$\text{By Vieta's Sums: } a+b+c=-S_{1}=0\text{ and } ab+bc+ca=-\frac{1}{2} \text{ and } abc=-S_{3}$

$\therefore P(x)=x^{3}-\frac{1}{2}x+S_{3}$

$\text{But } P'(x)=3x^{2}-\frac{1}{2}$

$\text{Critical points of this function are at: } x=\frac{+}{-} \frac{1}{\sqrt{6}}$

$\text{for the function to have three REAL roots, the } \textbf{maxima and minima } \text{of the function MUST BE } \textbf{greater than and less than zero respectively}$

$\text{GRAPHICALLY: This can be easily understood by the fact that for the function to have three real roots, the }$

$\text{curve must cut the X-axis at three points, which is possible only if the curve BULGES OUTWARD of the axis,}$

$\text{which again, is only possible if the critical points lie alternately above and below the X axis for max. and min.}$

$\therefore P(\frac{1}{\sqrt{6}})>0 \implies S_{3}>\frac{1}{3\sqrt{6}} \implies abc<-\frac{1}{3\sqrt{6}}$

$\therefore (9abc)^{2}<\boxed{1.5}$

here a+b+c=0 and a^2+b^2+c^2=1 that's why we get from (a+b+c)^2 formula that ab+bc+ca= -1/2 and then a(b+c)+bc= -1/2 => a(-a)+ X/a= -1/2 [ let abc= X ] then X= a^3- a/2 then differentiated by a parameter for getting maximum value of X (abc) then we get a= +- Sq. root (1/6) and put this value above equation and get (9X)^2= (9abc)^2= 3/2 or 1.5

First let's semplify a bit the system of equation:

$\begin{cases} a+b+c=0 \\ a^{2}+b^{2}+c^{2}=1 \end{cases}$

$\begin{cases} c=-(a+b) \\ a^{2}+b^{2}+[-(a+b)]^{2}=1 \end{cases}$

$\begin{cases} c=-(a+b) \\ a^{2}+b^{2}+a^{2}+2ab+b^{2}=1 \end{cases}$

$\begin{cases} c=-(a+b) \\ 2a^{2}+2b^{2}+2ab=1 \end{cases}$

$\begin{cases} c=-(a+b) \\ a^{2}+b^{2}+ab=\frac{1}{2} \end{cases}$

Second semplify what we are searching for:

$(9abc)^{2} = 3^{4}a^{2}b^{2}c^{2} = 3^{4}a^{2}b^{2}[-(a+b)]^{2} = 3^{4}(ab)^{2}(a+b)^{2}$

Now we know from the equation $a^{2}+b^{2}+ab=\frac{1}{2}$ that if $a$ increase $b$ mast decrease to keep true the equation. Also knowing that the product of $ab$ is greater smoller is the difference between $a$ and $b$ , see if $a=b$ (a-b=0, the smollest possible) is a solution and, if it is, what is the result. So be $x=a=b$ :

$x^{2}+x^{2}+x\times x=\frac{1}{2}$

$3x^{2}=\frac{1}{2}$

$x^{2}=\frac{1}{6}$

And finally find (9abc)^{2}:

$3^{4}(ab)^{2}(a+b)^{2} = 3^{4}(x\times x)^{2}(2x)^{2} = 3^{4} (x^{2})^2 4x^{2} = 3^{4}\times (\frac{1}{6})^{2}\times \frac{4}{6} = \frac{3}{2} = 1.5$

So the solution is $\boxed{1.5}$

$y + z = -x$

$y^2 + z^2 = 1 - x^2$

apply CS inequality we get

root(2/3) >= x

so let x = root(2/3)

then $y + z$ = -root(2/3)

and

$y^2 + z^2 = 1/3$

so y = z = -root2/2.root3

solving gives answer 1.5

I initially tried to solve this problem using a Cauchy-Schwarz inequality. There must be an error in my approach, because it does not yield the correct solution. I was wondering if someone could take a look at my approach and identify my error. I am rather new to CS inequalities, so I am not always sure when they are applicable or what kind of assumptions need to be made.

I obtained the following identity using a similar approach to others: $ab+ac+bc=-\frac{1}{2}$

I then obtained a new identity by squaring both sides of this equation:

$a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2=\frac{1}{4}$

$a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c)=\frac{1}{4}$

$a^2b^2+a^2c^2+b^2c^2+2abc(0)=\frac{1}{4}$

$a^2b^2+a^2c^2+b^2c^2=\frac{1}{4}$

I then set up the Cauchy-Schwarz inequality like so:

$(a^2b^2+a^2c^2+b^2c^2)(c^2+b^2+a^2)\ge(abc+abc+abc)^2$

$(\frac{1}{4})(1)\ge(3abc)^2$

$\frac{1}{4}\ge(3abc)^2$

Multiplying both sides of the inequality by $9$ yields:

$\frac{9}{4}\ge(9abc)^2$

This implies that the maximum value of $(9abc)^2$ is $\frac{9}{4}$ , which is not true.

From a²+b²+c²=1 We can replace a²+b²+c²= sin²(x)×(cos²(y)+sin²(y))+cos²(y) i.e. a=sin(x)sin(y),b=cos(x)sin(y),c=cos(y) So because a+b+c=0 we have sin(y)(sin(x)+cos(x))+cos(y)=0 I.e. sin(x)+cos(x)=-cos(y)/sin(y) So squaring that we get 1+sin(2x)=(1-sin²(y))/sin²(y) so sin(2x)=1/sin²(y)-2=(1-2sin²(y))/sin²(y) So 81a²b²c²=81sin²(x)cos²(x)sin⁴(y)cos²(y) =81/4×sin²(2x)sin⁴(y)(1-sin²(y))= 81/4×(1-2sin²(y))²(1-sin²(y)) Put sin²(y)=p, 0<=p<=1 So we need to find max of f(p)=81/4×(1-2p)²(1-p) i.e. f(p)=81/4×(-4p³+8p²-5p+1) So 0=f'(p)=81/4(-12p²+16p-5) p=1/2 or p=5/6 So we choose p=5/6 because f(0.5)=0 So max is 81/4×(1-10/6)²(1-5/6)= 81/4×4/9×1/6=3/2=1.5

My solution is quite intuitive though. Please correct me if I make some mistakes.

First notice that if one of them is 0 then the product is 0. If only one of them is negative and the rest are positive then the product is negative. Similarly, if only one of them is positive and the rest are negative then the product is positive.

Therefore, WLOG assume a,b < 0 and c > 0 $c = -a - b$ $c^2 = (a+b)^2$ Thus, $a^2 + b^2 + c^2 = 2(a^2+b^2+ab) = 1$ So, now we are maximizing $3^4a^2b^2(a^2+b^2+2ab) = 3^4a^2b^2(a^2+b^2+ab + ab) =3^4a^2b^2(1/2 + ab)$

Let X = ab >0, so we are maximizing $x^2(1/2+x)$ which is an increasing function on X. That is, if we can maximize ab we will have the answer to this. By AM-GM inequality (note that AM-GM assume all values are positive but in our case they are all negative anyways, so we may assume a' = |a| >0 and everything work just fine) $1/6 = (a^2+b^2+ab)/3 \geq ab$ Therefore, the maximum occurs at a=b (x = 1/6) Thus

$(9abc)^2 \leq 3^4(1/6)^2(1/2+1/6) = 3/2$

EDIT: I forgot the rule of thumb :D ... have to make sure those values exist. This is trivial: by AM-GM we know that a=b makes this work. That is |a|= |b| = 1/sqrt(6) (ie., a =b = -1/sqrt(6), c = 2/sqrt(6))

Lets proceed by trigonometry as all a,b,c have upper and lower limits. Solving for 'a' in terms of 'b' after eliminating 'c' using these two equations we get -sqrt(2/3)<= b sqrt(2/3). ( just make discriminant so obtained positive as 'a' is real.) so let; b=sqrt(2/3)(sin x). putitng this b back in 'a' and 'c' equation gives .................. 81aabbcc = (3/2)(sin x)(sin x) { (sin x)(sin x) - 3(cos x)(cos x) }. Clearly this maximum when sin x =1. which gives the maximum value as.......................1.5.