Classic Inequality

Algebra Level 4

If a , b , c , d a,b,c,d are four distinct positive real numbers and if 3 s = a + b + c + d 3s=a+b+c+d , then a b c d ( s a ) ( s b ) ( s c ) ( s d ) \dfrac{abcd}{(s-a)(s-b)(s-c)(s-d)} must be larger than:


The answer is 81.

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1 solution

Applying the AM-GM inequality, we have:

( s a ) ( s b ) ( s c ) ( s a + s b + s c 3 ) 3 = d 3 27 (s-a)(s-b)(s-c)\le\left(\dfrac{s-a+s-b+s-c}{3}\right)^3=\dfrac{d^3}{27}

Similarly, we have ( s b ) ( s c ) ( s d ) a 3 27 ; ( s c ) ( s d ) ( s a ) b 3 27 ; ( s d ) ( s a ) ( s b ) c 3 27 (s-b)(s-c)(s-d)\le\dfrac{a^3}{27};(s-c)(s-d)(s-a)\le\dfrac{b^3}{27};(s-d)(s-a)(s-b)\le\dfrac{c^3}{27}

Hence, ( s a ) 3 ( s b ) 3 ( s c ) 3 ( s d ) 3 a 3 b 3 c 3 d 3 3 12 (s-a)^3(s-b)^3(s-c)^3(s-d)^3\le\dfrac{a^3b^3c^3d^3}{3^{12}} or a b c d ( s a ) ( s b ) ( s c ) ( s d ) 81 \dfrac{abcd}{(s-a)(s-b)(s-c)(s-d)}\ge 81

So, the minimum value of a b c d ( s a ) ( s b ) ( s c ) ( s d ) \dfrac{abcd}{(s-a)(s-b)(s-c)(s-d)} is 81 \boxed{81} .

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