1if u used AM-GM the it wud be
{(x2)3/2 + (y2)3/2 + (z2)3/2 }/3 ≥ xyz
A frnd of mine gave me this sum to solve but i am not being able to do so...(cant match the ans)... the q is:
x2+y2+z2=27
then x3+y3+z3 has:
a) min value 81
b)max value 81
c)min value 27
d) max value 27
ans: b)
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8 Answers
1I got a) this is how i solved [ (x2)3/2+(y2)3/2 + (z2)3/2]/3 >= [{x2+y2+z2}/3]3/2
and hence got a)..
tell me where did i go wrong?
1
1the ans is A
jus a check put x=0
y=0
z=3√3
and the lower eq givs 140 around so ans is not b ....a) can b easily obtanied by takin x=y=z
341Power mean inequality:
\left(\frac{x^3+y^3+z^3}{3} \right)^{\frac{1}{3}} \ge \left(\frac{x^2+y^2+z^2}{3} \right)^{\frac{1}{2}}
It immediately follows that x^3+y^3+z^3 \ge 81
