### 2 Answers

- kbLv 75 years agoFavourite answer
The main idea is to repeatedly use the AM-GM (Arithmetic Mean/Geometric Mean Inequality): (x + y)/2 ≥ √(xy) for any real x, y ≥ 0.

[This is true, being equivalent to (x - y)² ≥ 0.]

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Let a, b, c be real numbers.

(i) First of all, we show that a²b² + a²c² + b²c² ≥ abc(a + b + c).

To this end:

(a² + b²)/2 ≥ √(a²b²), via AM-GM

..................= |ab|

..................≥ ab.

Hence, (a² + b²)/2 ≥ ab.

==> c²(a² + b²)/2 ≥ abc².

Similarly, a²(b² + c²)/2 ≥ a²bc and b²(c² + a²)/2 ≥ ab²c.

Adding these three inequalities together yields

a²b² + a²c² + b²c² ≥ abc² + a²bc + ab²c = abc(a + b + c).

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(ii) Next, we show that a⁴ + b⁴ + c⁴ ≥ a²b² + a²c² + b²c².

(a⁴ + b⁴)/2 ≥ √(a⁴b⁴), via AM-GM

..................= a²b².

Similarly, (b⁴ + c⁴)/2 ≥ b²c² and (a⁴ + c⁴)/2 ≥ a²c².

Adding these three inequalities together yields

a⁴ + b⁴ + c⁴ ≥ a²b² + a²c² + b²c²

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Finally, we use (i) and (ii) to establish the inequality in question.

a⁴ + b⁴ + c⁴ ≥ a²b² + a²c² + b²c² ≥ abc(a + b + c).

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I hope this helps!

- JOHNLv 75 years ago
Here is an alternative solution to Kb's:

WLOG, assume a ≤ b ≤ c. Apply the rearrangement inequality to the 4 sets {a, b, c}, {a, b, c}, {b, c, a}, {c, a, b}. We have:

a⁴ + b⁴ + c⁴ = a x a x a x a + b x b x b x b + c x c x c x c ≥ a x a x b x c + b x b x c x a + c x c x a x b = a²bc + b²ca + c²ab = abc(a + b + c).