A modification

Algebra Level 4

If non-negative real numbers x x , y y and z z , each ≥ 1 4 \ge \frac 14 , are such that: x + y + z = 1 x+y+z=1 . Find the minimum value of the expression below:

x 3 + y 3 + z 3 + 3 x y z \large x^3+y^3+z^3+3xyz

If your answer is j j , enter the answer as 450 j 450j .


The answer is 100.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Using the identities:

1. x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) 1. x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)

2. ( x + y + z ) 2 = x 2 + y 2 + z 2 + 2 x y + 2 y z + 2 z x 2. (x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx

The given inequality can be rewritten as :

2 ( x 3 + y 3 + z 3 ) − 3 2 ( x 2 + y 2 + z 2 ) + 1 2 ≥ 2 9 2(x^3+y^3+z^3) - \frac{3}{2}(x^2+y^2+z^2) +\frac{1}{2} \ge \frac{2}{9}

Consider the following function:

f ( t ) = 2 t 3 − 3 2 t 2 + 1 6 f(t) = 2t^3 - \frac{3}{2}t^2 + \frac{1}{6}

The above function becomes convex for t ≥ 1 4 \ge \frac{1}{4}

So, using Jensen's inequality on a , b a,b and c c

f ( a ) + f ( b ) + f ( c ) 3 ≥ f ( 1 3 ) \frac{f(a)+f(b)+f(c)}{3} \ge f(\frac{1}{3})

We get,

2 ( x 3 + y 3 + z 3 ) − 3 2 ( x 2 + y 2 + z 2 ) + 1 2 ≥ 2 / 9 2(x^3+y^3+z^3) - \frac{3}{2}(x^2+y^2+z^2) +\frac{1}{2} \ge 2/9

Fact \textbf{Fact} : By using some identities it can be shown that the given inequality is equivalent to an IMO 1984 problem

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Is my method correct?? I found the minimum of x^3+y^3+z^3 by Power Mean inequality and the point where the minimum was attained, I substituted those values in 3xyz.

Aaghaz Mahajan - 3 years, 7 months ago

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@Aaghaz Mahajan your method is wrong as the points where x 3 + y 3 + z 3 x^3+y^3+z^3 attains its minimum may not be the points where x y z xyz also attains its minimum.

A Former Brilliant Member - 3 years, 7 months ago

nice question and nice solution !(upvoted)

Rishu Jaar - 3 years, 7 months ago

Does anyone understand (and is willing to explain) how this former brilliant member got the expression 2 ( x 3 + y 3 + z 3 ) − 3 2 ( x 2 + y 2 + z 2 ) + 1 2 2(x^3+y^3+z^3)-\frac{3}{2}(x^2+y^2+z^2)+\frac{1}{2} ? I tried to reproduce it but failed.

James Wilson - 5 months ago

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