$\text{LHS}-\text{RHS}=\frac14(a+b)(a-b)^2$

Let $a, b\geq 0 .$ Prove that $$(a + b)^3\ge\frac{3\sqrt 3}{2}( a^2b + 2ab^2)$$$$(a + b)^3\ge\frac{2(7\sqrt 7-10)}{9}( a^2b + 3ab^2)$$$$(a + b)^3\ge a^3 + ab^2$$$$(a + 2b)^3\ge 6\sqrt 3( a^2b + 2ab^2)$$$$(a + 3b)^3\ge (14\sqrt 7-20)( a^2b + 2ab^2)$$

Let $a, b,c\geq 0 .$ Prove that $$(a+b+c)^3\geq\frac{27}{4}(a^2b+b^2c+c^2a)$$$$(a+2b+c)^3\geq\frac{27}{4}(a^2b+b^2c+c^2a)$$$$(a+2b+2c)^3\geq\frac{27}{2}(a^2b+b^2c+c^2a)$$$$(a+3b+3c)^3\geq\frac{81}{4}(a^2b+b^2c+c^2a)$$$$(a+4b+4c)^3\geq 27(a^2b+b^2c+c^2a)$$$$(a+b+c)^3\geq\frac{27}{8}(a^2b+2b^2c+c^2a)$$$$(a+b+c)^3\geq\frac{9}{4}(a^2b+3b^2c+c^2a)$$$$(a+b+c)^3\geq\frac{27}{16}(a^2b+4b^2c+c^2a)$$here

The second follows from the first as $a+b+c\le a+2b+c$.

Thanks.

sqing wrote: Let $a, b,c\geq 0 .$ Prove that $$(a+b+c)^3\geq\frac{27}{4}(a^2b+b^2c+c^2a)$$]

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sqing wrote: Let $a, b\geq 0 .$ Prove that $$(a + b)^3\ge\frac{3\sqrt 3}{2}( a^2b + 2ab^2)$$$$(a + b)^3\ge\frac{2(7\sqrt 7-10)}{9}( a^2b + 3ab^2)$$$$(a + b)^3\ge a^3 + ab^2$$$$(a + 2b)^3\ge 6\sqrt 3( a^2b + 2ab^2)$$$$(a + 3b)^3\ge (14\sqrt 7-20)( a^2b + 2ab^2)$$

Attachments:

Very nice

parmenides51 wrote: Prove that for all positive real numbers $a$ and $ b$: $$\frac{(a + b)^3}{4} \ge a^2b + ab^2$$ For positive real numbers $a$ and $ b$, it is true that $$(a+b)(a-b)^2 \geq 0$$$$a^3-a^2b-ab^2+b^3 \geq 0$$$$a^3+3a^2b+3ab^2+b^3 \geq 4a^2b + 4ab^2$$$$(a+b)^3 \geq 4a^2b+4ab^2$$$$\frac{(a+b)^3}{4} \geq a^2b+ab^2$$

For positive real numbers $a$ and $ b$, $$3a^2+2ab+b^2 \geq 0$$$$3a^2b+2ab^2+b^3 \geq 0$$$$a^3+3a^2b+3ab^2+b^3 \geq a^3+ab^2$$$$(a+b)^3 \geq a^3+ab^2$$