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Prove the inequality (a^3-c^3)/3≥abc((a-b)/c+(b-c)/a)

lfdahl

Well-known member
Nov 26, 2013
719
Let $a, b$ and $c$ be non-zero real numbers, and let $a\ge b \ge c$. Prove the inequality:

$$\frac{a^3-c^3}{3} \ge abc\left(\frac{a-b}{c}+\frac{b-c}{a}\right)$$

When does equality hold?

Source: Nordic Math. Contest
 

kaliprasad

Well-known member
Mar 31, 2013
1,279
Let $a, b$ and $c$ be non-zero real numbers, and let $a\ge b \ge c$. Prove the inequality:

$$\frac{a^3-c^3}{3} \ge abc\left(\frac{a-b}{c}+\frac{b-c}{a}\right)$$

When does equality hold?

Source: Nordic Math. Contest

because $a >= b$ so $(a-b)>0$ or $(a-b)^3>= 0$ or $a^3-3a^2b+3ab^2-b^3>=0$
or $a^3-b^3 >= 3ab(a-b)$
similarly $b^3-c^3 >=3bc(b-c)$
adding we $a^3 -c^3 >= 3abc(\frac{a-b}{c}+\frac{b-c}{a})$
dividing both sides by 3 we get the result

they are equal if a=b =c


rationale


equal if $(a-b)^3 + (b-c)^3 = 0$ sum of 2 non negative numbers when each is zero

 
Last edited:

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,120
So simple.... (Bow)

-Dan
 

lfdahl

Well-known member
Nov 26, 2013
719