The following inequality is also true

luofangxiang wrote: The following inequality is also true $4(x^3+y^3)\geq(x+y)^3$ and AM-GM!

arqady wrote: luofangxiang wrote: The following inequality is also true $4(x^3+y^3)\geq(x+y)^3$ and AM-GM! Can you post a solution?

$$(a+b)^3+(c+d)^3+8\geq\frac{(a+b+c+d)^3}{4}+8=2\cdot\frac{(a+b+c+d)^3}{8}+8\geq$$$$\geq3\sqrt[3]{\left(\frac{(a+b+c+d)^3}{8}\right)^2\cdot8}=\frac{3}{2}(a+b+c+d)^2\geq\sum_{sym}ab.$$

$$(a+b)^3+(c+d)^3+2(a^2+b^2+c^2+d^2) \ge \frac{(a+b)^3}{1^2}+\frac{(c+d)^3}{1^2}+8 \ge \frac{(a+b+c+d)^3}{(1+1)^2}+8=\frac{(a+b+c+d)^3}{4}+8$$by the Engel form of Holder's ineq. Now, $$\frac{(a+b+c+d)^3}{4}+8 = \frac{(a+b+c+d)^3}{8}+\frac{(a+b+c+d)^3}{8}+8 \ge \frac{3}{2}(a+b+c+d)^2$$$$\frac{3}{2}(a+b+c+d)^2 = \frac{3}{2}(a^2+b^2+c^2+d^2+2 \sum_{sym} ab ) = \frac{3}{2}(a^2+b^2+c^2+d^2)+3 \sum_{sym} ab$$We only need to prove $$\frac{3}{2}(a^2+b^2+c^2+d^2) \ge \sum_{sym} ab$$$$\iff 3(a^2+b^2+c^2+d^2) \ge 2 \sum_{sym} ab$$which is obviously true because $$\sum_{sym} (a^2+b^2) \ge \sum_{sym} (2ab) = 2 \sum_{sym} ab$$

$$(a+b)^3+(c+d)^3 \ge (S^3)/4$$and we define $S=a+b+c+d$ and we can also prove: $ \frac{3S^2}{2} \ge\ 4(ab+bc+cd+da+ac+db)$ so we just have to prove $ \frac{S^3}{4} +8 \ge\ \frac{3S^2}{2}$ which is obvious owing to the fact $\frac{S^3}{4} +8 - \frac{3S^2}{2} =\frac{S^3+32-6S^2}{4}=\frac{(S-4)^2(S+2)}{4}$

Let $x=a+b,y=c+d$ the original inequality rewrites as $x^3+y^3-4xy+2(a-b)^2+2(c-d)^2$ And $x+y\ge \sqrt{\sum a^2}\ge 2$ so we are done by AM-GM.

It's easy to check that $\sum{a} \leq \sum{a^2}$ $4\sum{ab} = 2 (\sum{a})^2 - 2 \sum{a^2} $ and $(a+b)^3+(c+d)^3 = (\sum{a})^3 - 3 (a+b)(c+d)\sum{a}$ and $(a+b)(c+d) \leq \frac{1}{4} (\sum{a})^2 $ $\leftrightarrow (\sum{a})^3 - 3(a+b)(c+d)\sum{a} +4\sum{a^2} -2(\sum{a})^2 \geq 0 \leftrightarrow (\sum{a})^2 - 3(a+b)(c+d) + 4 \frac{\sum{a^2}}{\sum{a}} - 2\sum{a} \geq 0$ $\leftrightarrow (\sum{a})^2 - \frac{3}{4} .(\sum{a})^2 +\frac{4\sum{a^2}}{\sum{a}} -2\sum{a} \geq 0 $ $\leftrightarrow (\frac{1}{2} \sum{a} -2 )^2 + 4\frac{\sum{a^2} - \sum{a}}{\sum{a}} \geq 0 $ and the problem is finished $\blacksquare$

Lemma : Let $x,y$ be positive real numbers $x^3 + y^3 \ge \frac{(x+y)^3}{4}$. Proof : We want to prove $4x^3 + 4y^3 \ge x^3 + 3x^2y + 3xy^2 + y^3$ or $x^3 + y^3 \ge x^2y + xy^2$ or $x^2(x-y) + y^2(y-x) \ge 0$ or $(x^2 - y^2)(x-y) \ge 0$ or $(x-y)^2(x+y) \ge 0$ which is true. $(a+b)^3+(c+d)^3+2(a^2+b^2+c^2+d^2) \ge \frac{(a+b+c+d)^2}{4} + 8 = 2\frac{(a+b+c+d)^2}{8} + 8 \ge 3\sqrt[3]{(\frac{(a+b+c+d)^3}{8})^2 . 8} = \frac{3}{2}(a+b+c+d)^2 = \frac{3}{2}(a^2+b^2+c^2+d^2)+3 \sum_{sym} ab$ so we need to prove $\frac{3}{2}(a^2+b^2+c^2+d^2) \ge \sum_{sym} ab$ or $3(a^2+b^2+c^2+d^2) \ge 2\sum_{sym} ab$ which is easy to prove with AM-GM .

This lemma is key ! : $x^3 + y^3 \ge \frac{(x+y)^3}{4}$ . The prove of this lemma is easy ! . by your self. $(a+b)^3+(c+d)^3+2(a^2+b^2+c^2+d^2) \ge 4(ab+bc+cd+da+ac+bd) \to \frac{(a+b+c+d)^3}{8}+ \frac{(a+b+c+d)^3}{8}+8 \geq \frac{3}{2}(a+b+c+d)^2$ Now we need prove : $3\sum a^2 +6\sum ab \geq 8 \sum ab \to (a-b)^2+(b-c)^2+(c-a)^2 +a^2+b^2+c^2 \geq 0$ . $\blacksquare$