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prove $$a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b),$$$a,b,c>0$

Obviously this is a direct consequence of the third degree schur's inequality.

I was wondering if this could be proved without this theorem ,or uvw but through basic methods like AM-GM,C-S etc.

Michael Rozenberg
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Albus Dumbledore
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  • Take any algebraic proof and look what it is doing step by step when writing the inequality in the form $$a(a-b)(a-c) +b(b-a)(b-c)+c(c-a)(c-b)\ge 0$$ and performing the calculus parallely. (Sure, this is the Schur idea.) For instance, starting with $a\le b\le c$ (and introducing new variables) we more or less use the same argument. – dan_fulea Sep 05 '20 at 18:15

6 Answers6

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Easiest proof is actually the proof of Schur. Alternatively, you can even prove it using simple calculus.

Assume $a\geq b\geq c.$ Consider:

$$f(x) = x^3-x^2(b+c) - x(b^2+c^2) +3xbc+b^3+c^3-bc(b+c),\,\, x\geq b$$ Then, $f'(x) = 3x^2-2x(b+c) - b^2-c^2+3bc$ and $f''(x) = 6x - 2b-2c\geq 0.$ So $f'$ is increasing on $[b,\infty)$ and as such: $$f'(x)\geq f'(b) = 3b^2 - 2b^2-2bc-b^2-c^2+3bc = c(b-c)\geq 0$$ which means $f$ is increasing on its domain. Finally,

$$f(a)\geq f(b) = b^3-b^3-b^2c-b^3-bc^2+3b^2c+b^3+c^3-b^2c-bc^2 = $$ $$ = c^3+b^2c-2bc^2 = c(c-b)^2\geq 0.$$

dezdichado
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  • i liked your solution!,However do you have any idea to solve without calculus using standard inequalities.Thanks for the effort – Albus Dumbledore Sep 05 '20 at 17:53
  • @Quantum I am unaware of an AM-GM proof and think that is impossible in the original form due to the term $abc$ on the LHS. Any attempt of AM-GM or muirhead fails directly because $(1,1,1)$ is majorized by any other triplet of $(\alpha, \beta, \gamma)$ such that $\alpha + \beta+\gamma = 3.$

    Subsituting $a = c+x+y$ and $b = c+x$, $x,y\geq 0$ will reduce the inequality to a quadratic inequality in $c$, which is not so much calculus as completing the square.

     – dezdichado Sep 05 '20 at 18:37
3

Actually, a solution by AM-GM is possible. Schur is equivalent to:

$$abc\geq (-a+b+c)(a-b+c)(a+b-c).$$

In the case that all the terms of RHS is non-negative, then applying AM-GM like $$\sqrt{(-a+b+c)(a-b+c)}\leq\dfrac{-a+b+c +a-b+c}{2}=c$$ will yield the inequality.

If exactly one or all three of the terms negative, then there is nothing to prove as the product will be negative. If say, $a-b+c$ and $-a+b+c$ were negative, then their sum $2c$ is negative, which contradicts the positivity assumption, so we are done.

dezdichado
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Let $a\geq b\geq c$.

Thus, $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}a(a-b)(a-c)\geq$$ $$\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0.$$

Michael Rozenberg
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1

Another way.

Let $a\geq b\geq c$.

Thus, $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}(a^3-abc-a^2b-a^2c+2abc)=$$ $$=\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{2}-c\right)=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b-c)\geq$$ $$\geq\frac{1}{2}((a-c)^2(a+c-b)+(b-c)^2(b+c-a))\geq$$ $$\geq\frac{1}{2}((b-c)^2(a-b)+(b-c)^2(b-a))=0.$$

Michael Rozenberg
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Suppose $a \geqslant b \geqslant c,$ we get $$a^3+b^3+c^3+3abc - \sum ab(a+b)$$ $$=(c^3+b^2c+ca^2-abc-ac^2-bc^2)+(a^3+b^3+4abc-a^2b-ab^2-2ca^2-2b^2c)$$ $$=(a^2+b^2+c^2-ab-bc-ca)c+(a+b-2c)(a-b)^2 \geqslant 0.$$

nguyenhuyenag
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1

We write your inequality as$:$

$$\sum a(a-b)(a-c) \geqslant 0.$$

Due to symmetric$,$ by assuming $a :\neq \max\,\{a,b,c\},$ we have$:$

$$\sum a(a-b)(a-c) = a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)$$

$$=a(a-b)(a-c) +[b(b-c)(b-a)+c(c-a)(c-b)]$$

$$=a(a-b)(a-c) +(b-c)^2(b+c-a) \geqslant 0,$$

which is SOS-Schur method.

Else method is$:$

$$(a+b+c) \sum a(a-b)(a-c) = \sum a^2(a-b)(a-c) +\sum (a^3b+b^3a-2a^2b^2)$$

which is just Schur degree $4$ and AM-GM.

Schur degree $4$ is easy prove.

NKellira
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