parmenides51 wrote: Let $a, b$ and $c$ be pairwise different natural numbers. Prove $\frac{a^3 + b^3 + c^3}{3} \ge abc + a + b + c$. When does equality holds? (Karl Czakler) $a^3+b^3+c^3 - 3abc = \frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$ We are given $|a-b| \geq 1,|b-c| \geq 1,|c-a| \geq 1$ Thus from triangle inequality, $|a-b| +|b-c| \leq |a-c| $ $\implies |c-a| \geq |a-b| +|b-c| \geq 1+1 =2$ $\implies |c-a| \geq 2$ Hence, $((a-b)^2+(b-c)^2+(c-a)^2) \geq 1 + 1 + 4 = 6$ $\implies \frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \geq \frac{1}{2}(a+b+c)(1 + 1 + 4) = 3(a+b+c)$ $\implies a^3+b^3+c^3 - 3abc \geq 3(a+b+c)$ $\implies \frac{a^3 + b^3 + c^3}{3} \geq abc + a + b + c$ Equality holds iff $|a-b|=|b-c| = 1$ $\implies a-b=b-c$ or $ a-b=c-b$ $\implies a+c=2b$ or $a=c$(--><--) So, equality holds iff a,b,c are in A.P. with common difference =1.

This is a very old problem.

parmenides51 wrote: Let $a, b$ and $c$ be pairwise different natural numbers. Prove $\frac{a^3 + b^3 + c^3}{3} \ge abc + a + b + c$. When does equality holds? (Karl Czakler) See, for example, here https://www.cut-the-knot.org/arithmetic/algebra/InequalityInIntegers2.shtml

sanyalarnab wrote: parmenides51 wrote: Let $a, b$ and $c$ be pairwise different natural numbers. Prove $\frac{a^3 + b^3 + c^3}{3} \ge abc + a + b + c$. When does equality holds? (Karl Czakler) $a^3+b^3+c^3 - 3abc = \frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$ We are given $|a-b| \geq 1,|b-c| \geq 1,|c-a| \geq 1$ Thus from triangle inequality, $|a-b| +|b-c| \leq |a-c| $ $\implies |c-a| \geq |a-b| +|b-c| \geq 1+1 =2$ $\implies |c-a| \geq 2$ Hence, $((a-b)^2+(b-c)^2+(c-a)^2) \geq 1 + 1 + 4 = 6$ $\implies \frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \geq \frac{1}{2}(a+b+c)(1 + 1 + 4) = 3(a+b+c)$ $\implies a^3+b^3+c^3 - 3abc \geq 3(a+b+c)$ $\implies \frac{a^3 + b^3 + c^3}{3} \geq abc + a + b + c$ Equality holds iff $|a-b|=|b-c| = 1$ $\implies a-b=b-c$ or $ a-b=c-b$ $\implies a+c=2b$ or $a=c$(--><--) So, equality holds iff a,b,c are in A.P. with common difference =1. Wrong soln. U have exactly reversed the triangle ineq

parmenides51 wrote: Let $a, b$ and $c$ be pairwise different natural numbers. Prove $\frac{a^3 + b^3 + c^3}{3} \ge abc + a + b + c$. When does equality holds? (Karl Czakler) Rewrite the inequality as \[\left(a^3+b^3+c^3-3abc\right)-3(a+b+c)\geq 0\]which can be rewritten as \[(a+b+c)\left((a-b)^2+(b-c)^2+(c-a)^2-6\right)\geq 0\]which is true, because \(a\), \(b\), \(c\) are distinct naturals. Equality holds if and only if \(a\), \(b\), \(c\) are consecutive positive integers.

$a,a+x,a+x+y$

$a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ so we need to prove $a^2+b^2+c^2-ab-bc-ca \ge 3$. Let $b = a+x$ and $c = a+y$ so we need to prove $x^2 + y^2 - xy \ge 3$. Note that $x^2 + y^2 - xy \ge xy$ so as $x,y$ are natural we only need to check the case $x = 1,y = 2$ cause other than that $xy \ge 3$. in case $x = 1,y = 2$ we have $x^2 + y^2 - xy \ge 3$ so we're Done.

Observe that $$ \frac{a^3+b^3+c^3}{3}-abc = \frac{1}{6}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \;.........\;(1) $$Since $a \neq b \neq c \neq a$, the smallest possible value of $\, (a-b)^2+(b-c)^2+(c-a)^2 \,$ is $\, 1^2+1^2+2^2=6 \,$ which is achieved if and only if $\, a,b,c \,$ are consecutive. From $(1)$, we get $$ \frac{a^3+b^3+c^3}{3}-abc \geq a+b+c \quad\Longrightarrow\quad \frac{a^3+b^3+c^3}{3}\geq abc+a+b+c. $$Equality holds iff $\, a,b,c \, $ are consecutive.

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