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Given $a,b$ and $c$ are positive real numbers, prove that $$a^5 + b^5 + c^5 \ge abc(ab+bc+ca).$$

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closed as off-topic by choco_addicted, user91500, Shailesh, Davide Giraudo, Smylic Jun 6 '17 at 9:45

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I assume you mean $\geq$. Hint: rewrite the LHS as $$\frac{a^5}5+\frac{a^5}5+\frac{a^5}5+\frac{a^5}5+\frac{a^5}5+\frac{b^5}5+\frac{b^5}5+\frac{b^5}5+\frac{b^5}5+\frac{b^5}5+\frac{c^5}5+\frac{c^5}5+\frac{c^5}5+\frac{c^5}5+\frac{c^5}5,$$ then rearrange the terms into three groups of five and apply AM-GM to each group.

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While most people would resort to using manipulation to ...subdue the problem. I find the calculus approach still appealing and no less "elegant" in its own way...so let's start.

Divide both sides by $a^5$, and put $x = \dfrac{b}{a}, y = \dfrac{c}{a}$, then the "original" statement becomes: $1+x^5+y^5 \ge xy(xy+x+y)$, with $x, y > 0$. To this end, consider the two variable function $f(x,y) = 1+x^5+y^5 - x^2y-x^2y^2 - xy^2$. Our aim is to show: $f(x,y) \ge 0$ by showing that $f_{\text{min}} = 0$. We proceed by finding the critical points: $f_x = 0 = f_y\implies 5x^4-2xy-2xy^2-y^2 = 0 = 5y^4-2xy-2yx^2-x^2$. We subtract the latter equation from the former, and factor: $(x-y)(5(x^3+x^2y+xy^2+y^3)+2xy+x+y)=0\implies x - y = 0\implies x = y$. Substituting these values into the system $f_x = 0 = f_y$ we have: $5x^4-2x^3-3x^2 = 0\implies x^2(5x+3)(x-1) = 0\implies x = 1$ since $x > 0$. Thus the only critical point is $(1,1)$. We have:$f_{xx}(1,1) = 16 > 0, f_{xy}(1,1) = -8, f_{yy}(1,1) = 16\implies D = (f_{xx}f_{yy} - f^2_{xy})|_{(1,1)} = 16^2 - 8^2 = 256-64 = 192 > 0$. Thus by the calculus' D-test, $(x,y) = (1,1)$ is the relative minima and since the domain is open in $\mathbb{R^2}$ ( the first quandrant ), this point is also the global minima which means $f_{\text{min}} = f(1,1) = 0\implies f(x,y) \ge 0\implies 1+x^5+y^5 \ge xy(xy+x+y)$ which is the desire claim we sought to show.

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$$\sum_{cyc}(a^5-a^2b^2c)=\frac{1}{2}\sum_{cyc}(2a^5-a^3b^2-a^3c^2+a^3b^2+a^3c^2-2a^2b^2c)=$$ $$=\frac{1}{2}\sum_{cyc}(a^5-a^3b^2-a^3b^3+b^3+c^2a^3+c^2b^3-c^2a^2b-c^2ab^2)=$$ $$=\frac{1}{2}\sum_{cyc}((a^3-b^3)(a^2-b^2)+c^2(a^2-b^2)(a-b))=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2((a^2+ab+b^2)(a+b)+c^2(a+b))=$$ $$=\sum_{cyc}(a-b)^2(a+b)(a^2+b^2+c^2+ab)\geq0.$$ Done!

We see that your inequality is true even for all reals $a$, $b$ and $c$ such that

$a+b\geq0$, $a+c\geq0$ and $b+c\geq0.$

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