Let $a = x+y$, $b = y+z$, $c = z+x$. Expanding and simplifying gives $2 y (3 x^2+3 x y+3 x z+y^2+3 y z+3 z^2) > 0$ which is clearly true.

But you do this: $a,b,c$ be sides of a triangle,prove that: \[a^3+b^3+3bca \geq c^3+8h_a(h_c-r)a\]

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joybangla wrote: Let $a,b,c$ be sides of a triangle. Show that $a^3+b^3+3abc>c^3$. Use $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$.

\begin{align*} a^3+b^3-c^3+3abc &=(a+b)^3-c^3+3abc-3ab(a+b)\\ &=(a+b)^3-c^3-3ab(a+b-c)\\ & =(a+b-c)(a^2+b^2+c^2+2ab+bc+ca)-3ab(a+b-c)\\ & =(a+b-c)(a^2+b^2+c^2-ab+bc+ca)\\ & > (a+b-c)(a^2+b^2+c^2-ab-bc-ca) > 0 \end{align*}

joybangla wrote: Let $a,b,c$ be sides of a triangle. Show that $a^3+b^3+3abc>c^3$. Let $a,b,c$ be sides of a triangle. Show that $a^2+b^2+\frac{4abc}{a+b+c}>c^2$.

spikerboy wrote: \begin{align*} a^3+b^3-c^3+3abc &=(a+b)^3-c^3+3abc-3ab(a+b)\\ &=(a+b)^3-c^3-3ab(a+b-c)\\ & =(a+b-c)(a^2+b^2+c^2+2ab+bc+ca)-3ab(a+b-c)\\ & =(a+b-c)(a^2+b^2+c^2-ab+bc+ca)\\ & > (a+b-c)(a^2+b^2+c^2-ab-bc-ca) > 0 \end{align*} Nice. \[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\]\[\iff a^3+b^3-c^3+3abc=(a+b-c)(a^2+b^2+c^2-ab+bc+ca) \]

rachitgoel wrote: joybangla wrote: Let $a,b,c$ be sides of a triangle. Show that $a^3+b^3+3abc>c^3$. $a+b> c\implies a^3+b^3+3ab(a+b)> c^3\implies a^3+b^3+3abc>c^3.$ That last step is wrong. You cannot surely say from your steps that whether $a^3+b^3+3abc>c^3$ or $c^3+a^3+b^3+3abc$. For all we know it could be $c^3>a^3+b^3+3abc$. I cannot give any counter example because the inequality is true but your process is not. Review that please.

Since $a+b>c$ then $a^2-ab+b^2 +3ab > c^2$ then $c(a^2-ab+b^2)+3abc > c^3$, but $a^3+b^3= (a+b)(a^2-ab+b^2) > c(a^2-ab+b^2)$ thus conclusion.