Classical Inequalities addicts can do this. Part 3

Algebra Level 5

Let a , b , c , d a,b,c,d be all real numbers such that a 3 + b 3 + c 3 + d 3 = 41 a^3 + b^3 + c^3 + d^3 = 41 and ϕ = a + b + c + d ≥ 0 \phi = a+b+c+d \ge 0 .

What is the minimum value of ∑ a , b , c , d ( ϕ − a ) 3 \displaystyle \sum_{ a, \ b, \ c, \ d} (\phi - a)^3 ?

For more problems like this, try answering this set .


The answer is -41.

Christian Daang by Christian Daang , 20, Philippines

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1 solution

( a + b + c ) 3 + ( b + c + d ) 3 + ( c + d + a ) 3 + ( d + a + b ) 3 \left(a+b+c\right)^3+\left(b+c+d\right)^3+\left(c+d+a\right)^3+\left(d+a+b\right)^3

≥ ( − a ) 3 + ( − b ) 3 + ( − c ) 3 + ( − d ) 3 \ge\left(-a\right)^3+\left(-b\right)^3+\left(-c\right)^3+\left(-d\right)^3

= − ( a 3 + b 3 + c 3 + d 3 ) =-\left(a^3+b^3+c^3+d^3\right)

= − 41 =-41

3 Helpful 0 Interesting 1 Brilliant 0 Confused

Nice solution!

Steven Jim - 4 years ago

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