Can you solve it?

Algebra Level 5

∣ a b ( a 2 − b 2 ) + b c ( b 2 − c 2 ) + c a ( c 2 − a 2 ) ∣ ≤ M ( a 2 + b 2 + c 2 ) 2 | ab(a^{2} - b^{2}) + bc(b^{2} - c^{2}) + ca(c^{2} - a^{2}) | \leq M(a^{2} + b^{2} + c^{2})^{2}

Determine the least real number M M such that the inequality above holds for all real numbers a , b a,b and c c .

Give your answer to 2 decimal places.


The answer is 0.39.

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Mihir Chakravarti by mihir Chakravarti , 22, India

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1 solution

The left hand side factors like this: ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) (a-b)(b-c)(c-a)(a+b+c)

So just let x = a − b , y = b − c , z = c − a , s = a + b + c x=a-b, y=b-c, z=c-a, s=a+b+c Then the inequality becomes ∣ x y z s ∣ ≤ M 9 ( x 2 + y 2 + z 2 + s 2 ) 2 |xyzs|\leq \frac{M}{9}(x^{2}+y^{2}+z^{2}+s^{2})^{2} With the property that x + y + z = 0 x+y+z=0

Two of x , y , z x,y,z have the same sign. Let x , y x,y be that two. Now putting their arithmetic mean instead of them makes the inequality stricter (l.h.s. gets bigger and r.h.s. gets smaller)

So assume that x = y x=y . Now z = − 2 x z=-2x and the inequality becomes: ∣ 2 x 3 s ∣ ≤ M 9 ( s 2 + 6 x 2 ) 2 |2x^{3}s|\leq \frac{M}{9}(s^{2}+6x^{2})^{2} Now using the AM-GM inequality for

s 2 , 2 x 2 , 2 x 2 , 2 x 2 s^{2},2x^{2},2x^{2},2x^{2} we get ( s 2 + 6 x 2 ) 2 ≥ 16 8 s 2 x 6 = 32 2 s x 3 (s^{2}+6x^{2})^{2}\geq 16\sqrt{8s^{2}x^{6}}=32\sqrt 2sx^{3}

So M = 9 16 2 = 0.39 M=\frac 9{16\sqrt 2} = 0.39 works.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

if a = -1, b = -0.375 ,c = 0.375 and m = 0.39, you get on the left side of the inequality 0.64453125 and on

the right side 0.640224609375. It volates the problems conditions. Answer for m is closer to 0.400 such

as .399???. I solved the problem using an IPad running Pythonista in conjunction with numerical

techniques.

Frank Petiprin - 3 years, 8 months ago

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It should be closer to 0.399.

Frank Petiprin - 3 years, 8 months ago

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