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# Inequality 19(Vasile Cirtoaje)

Problem:

If $\displaystyle a,b,c$ are real numbers prove that $\displaystyle (a^2+b^2+c^2)^2\geq 3(a^3b+b^3c+c^3a)$.

Solution:

We are going to use the following well-known inequality:

$\displaystyle (x+y+z)^2\geq 3(xy+yz+zx)$.

So, if we transform the $\displaystyle x,y,z$ to $\displaystyle a^2+bc-ab,b^2+ca-bc,c^2+ab-ca$ respectively we have that

$\displaystyle \left(a^2+b^2+c^2\right)^{2}\geq 3\sum_{cyc}(a^2+bc-ab)(b^2+ca-bc)=3\sum_{cyc}a^3b=3(a^3b+b^3c+c^3a)$, Q.E.D.