AM - GM Inequality

$x^3+y^3+z^3-3xyz = (x+y+x)(x^2+y^2+z^2-xy-yz-zx)=1$

$x^2+y^2+z^2-xy-yz-zx=\dfrac12\left((x-y)^2+(y-z)^2+(z-x)^2\right) > 0 \Rightarrow x+y+z >0$

Therefore , we can apply AM - GM Inequality on $x+y+z$

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Call $x+y+z=A , x^2+y^2+z^2=B$

From the equation , we have $A\left(B-\dfrac{A^2-B}2\right)=1$

$\Rightarrow A(2B-A^2+B)=2$

$\Rightarrow 3AB-A^3=2$

$\Rightarrow 3AB=A^3+2 \quad \quad \cdots (1)$

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By AM - GM Inequality ,

$A^3+1+1 \geq 3A \quad \quad \cdots (2)$

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Combining $(1) , (2)$ ,

$B \geq 1$

Equality holds when $\boxed{x+y+z=x^2+y^2+z^2=1}$

This can be interpreted as a plane cutting a sphere, and so the locus is a circle. Explicitly, we have solutions like $(1, 0, 0)$ .

I will exploit the following identities:

1. x³ + y³ + z³ = (x+y+z)(x² + y²+ z² -xy - yz -zx)

2. (x+y+z)² = x² + y² + z² + 2(xy + yz + zx)

Let x² + y² + z² = l

And x + y + z = m

So the given expression is equivalent to :

m {l - (m² - l)/2} = 1

On simplifying we get

m³ - 3lm + 2 = 0

So , 3l = m² + 2/m

Now we need to minimise l or u can say 3l

Differentiating the above equation wrt to 'm'. it can be easily seen that minima is achieved at :

m = 1

When m = 1 , l = 1

So, the minimum value of l is 1

So x² + y² + z² <= 1

Try my problem ' Die '.