In fact $a^3+b^3+c^3+\frac{15}{4}abc \ge \frac14(a+b+c)^3$ too. Indeed \[ 4(a^3+b^3+c^3+\tfrac{15}{4}abc)-(a+b+c)^3 = 3(a^3+b^3+c^3+3abc - a^2(b+c) - b^2(c+a) - c^2(a+b)) \]which is nonnegative by Schur.

SD3 theorem

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Tumon2001, to be understandable better I post SD3 and CD3 theorem with proof.

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bitrak wrote: SD3 theorem bitrak： SD3 theorem,link?

bitrak wrote: Tumon2001, to be understandable better I post SD3 and CD3 theorem with proof. bitrak： Your picture if reply: https://artofproblemsolving.com/community/c6t243f6h1505857_arqadyyou_use_uvw_prove_inequality_with_xyz

Tumon2001 wrote: Let $a $, $b $, and $c $ be non-negative reals. Prove that $a^3+b^3+c^3+6abc\ge \frac{(a+b+c)^3}{4} $. Let $a,b,c,d>0$,prove \[27\,({a}^{4}+{c}^{4}+{b}^{4}+{d}^{4})+148\,abcd\geq \left( a+b+ c+d \right) ^{4}\]

5-var: $a,b,c,d,e>0$,prove \[{a}^{5}+{b}^{5}+{c}^{5}+{d}^{5}+{e}^{5}+{\frac {1845}{256}}\,abcde\geq { \frac {1}{256}}\, \left( a+b+c+d+e \right) ^{5}\] $6$-var $a,b,c,d,e,f>0$,prove \[{a}^{6}+{b}^{6}+{c}^{6}+{d}^{6}+{e}^{6}+{f}^{6}+{\frac {27906}{3125}} \,abcdef\geq {\frac {1}{3125}}\, \left( a+b+c+d+e+f \right) ^{6}\]

$n$-var \[\sum_{i=1}^n{x_i^n}-n\prod_{i=1}^n{x_i}\geq \frac{1}{(n-1)^{n-1}}((\sum_{i=1}^n{x_i})^n-n^n\prod_{i=1}^n{x_i})\]

Suppose that $a=max{{a,b,c}}$ then $2a \geq b+c$ then $LHS \geq a^3+b^3+c^3+3bc(b+c)=a^3+(b+c)^3 \geq RHS$

xzlbq wrote: bitrak wrote: SD3 theorem bitrak： SD3 theorem,link? All about SD3 and CD3 is in the picture above. Proved. Nothing else more.

If $a,b,c$ are arbitrary nonnegative real numbers, prove the inequality $$a^3+b^3+c^3+6abc\ge\frac14(a+b+c)^3$$with equality if and only if two of the numbers are equal and the third one equals zero.

Tumon2001 wrote: Let $a $, $b $, and $c $ be non-negative reals. Prove that $a^3+b^3+c^3+6abc\ge \frac{(a+b+c)^3}{4} $. Problem 212 Problem 444 New 87

Tumon2001 wrote: Let $a $, $b $, and $c $ be non-negative reals. Prove that $a^3+b^3+c^3+6abc\ge \frac{(a+b+c)^3}{4} $. Schur, of course

Let $ a,b,c\geq 0 $ and $ a +2b^2 + c = 1.$ Prove that $$ a^3 + b^3 + c^3 + 2abc \geq \frac {7}{32}$$