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Algebra Difficulty 3.3 AMC 10/12 Prove it North Macedonia

Let aa, bb, c∈Rc \in \mathbb{R} are such that abc=1abc = 1. Prove the inequality
a4+b4+c4≥a+b+c. a^4 + b^4 + c^4 \geq a + b + c.

Solution

By the trivial inequality (x−y)2+(y−z)2+(z−x)2≥0(x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0 we have that for every xx, yy, z∈Rz \in \mathbb{R} it holds
x2+y2+z2≥xy+yz+zx. x^2 + y^2 + z^2 \geq xy + yz + zx.
By applying this inequality, we get
a4+b4+c4≥a2b2+b2c2+c2a2=(ab)2+(bc)2+(ca)2≥(ab)(bc)+(bc)(ca)+(ca)(ab)=abc(a+b+c)=a+b+c \begin{align*} a^4 + b^4 + c^4 &\geq a^2 b^2 + b^2 c^2 + c^2 a^2 = (ab)^2 + (bc)^2 + (ca)^2 \\ &\geq (ab)(bc) + (bc)(ca) + (ca)(ab) = abc(a+b+c) = a+b+c \end{align*}
which finishes the proof. Equality holds if and only if a=b=c=1a = b = c = 1.

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