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#### lfdahl

##### Well-known member

- Nov 26, 2013

- 719

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

When does equality hold?

Source: Nordic Math. Contest

- Thread starter lfdahl
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- Thread starter
- #1

- Nov 26, 2013

- 719

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

When does equality hold?

Source: Nordic Math. Contest

- Mar 31, 2013

- 1,279

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

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:

- Aug 30, 2012

- 1,120

So simple....

-Dan

-Dan

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- #4

- Nov 26, 2013

- 719

Thankyou,
kaliprasad
for a clever solution.