JOHN
Lv 7
JOHN asked in Science & MathematicsMathematics · 5 years ago

For the real numbers a, b, c prove that a⁴ + b⁴ + c⁴ ≥ abc(a + b + c).?

2 Answers

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  • kb
    Lv 7
    5 years ago
    Favourite 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.]

    ----

    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).

    ----

    (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

    ----

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

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

    -------

    I hope this helps!

  • JOHN
    Lv 7
    5 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).

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