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$$4(a+b+c)^3\ge 27(a^2b+b^2c+c^2a)$$, $a,b,c\ge0$

I tried to work out this inequality the way I did So we have to show $$ 4(a + b + c) ^ 3 \ge 27 (a ^ 2b + b ^ 2c + c ^ 2a + abc) $$

One way is to use cyclic symmetry, and WLOG assumes $ a $ is $ a, b, c $ minute. Then we can write $ b = a + x, c = a + y $, where $ x, y \ge 0 $. Now the inequality reduces to 9 $$ (x ^ 2-xy + y ^ 2) + (x-2y) ^ 2 (4x + y) \ge 0 $$ which is obvious. Also from the above, we get then equality is possible iff $ x = y = 0 $ or when $ a = 0, x = 2y $, i.e. when $ (a, b, c) = (1, 1, 1) $ or permutation $ (0, 2, 1) $.

I think it's a mistake please help

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  • $\begingroup$ There is something off in the last inequality, all the terms are not homogeneous. Nonetheless, you're mostly right, if you can prove something that even better, that's mean you've done a great job. $\endgroup$
    – Paresseux Nguyen
    Dec 4, 2020 at 13:31

1 Answer 1

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The last inequality should be $$9(x^2-xy+y^2)a+4x^3-15x^2y+12xy^2+4y^3 \geqslant 0,$$ or $$9(x^2-xy+y^2)a+(4x+y)(x-2y)^2 \geqslant 0.$$ It's called BW method.

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