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prove that $$x^4 + y^4 + z^4 \geq xyz(x+y+z)$$

This AM-GM inequalities are seriously stumping me. I'd appreciate a full proof and explanation and hints for proving other inequalities like this. Thanks.

danimal
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user346756
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3 Answers3

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Use: $$a^2+b^2+c^2\ge ab+bc+ca$$ Proof:$$2(a^2+b^2+c^2)\ge 2(ab+bc+ca) \Leftrightarrow$$ $$\Leftrightarrow(a-b)^2+(b-c)^2+(c-a)^2\ge0$$

Then $$x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xy^2z+x^2yz+xyz^2=$$ $$=xyz(x+y+z)$$

Roman83
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AM–GM is invoked in two steps as follows:

$\begin{array}{rcl}x^4+y^4+z^4 &=& \dfrac{x^4+y^4}2+\dfrac{y^4+z^4}2+\dfrac{z^4+x^4}2 \\\\ &\ge& x^2y^2+y^2z^2+z^2x^2 \\\\ &=& \dfrac{x^2y^2+y^2z^2}2+\dfrac{y^2z^2+z^2x^2}2+\dfrac{z^2x^2+x^2y^2}2 \\\\ &\ge& xy^2z+yz^2x+zx^2y \\\\ &=& xyz(x+y+z)\end{array}$

George Law
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The inequality is equivalent to ( by expanding RHS) $$ x^4 + y^4 +z^4 \ge x^2yz + y^2xz + z^2xy$$ By AM-GM

$$ x^4+x^4 +y^4 +z^4 \ge 4(x^8y^4z^4)^\frac{1}{4} = 4x^2yz$$ $$ y^4+y^4 +x^4 +z^4 \ge 4(y^8x^4z^4)^\frac{1}{4} = 4y^2xz$$ $$ z^4+z^4 +x^4 +y^4 \ge 4(z^8x^4y^4)^\frac{1}{4} = 4z^2xy$$ Adding them up we get

$$4(x^4 +y^4 +z^4) \ge 4(x^2yz+y^2xz + z^2xy)$$ $$ x^4 +y^4 +z^4 \ge x^2yz+y^2xz + z^2xy $$

Ehit Karim
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