given that a2 + b2+ c2 = 8
prove that a3 + b3 + c 3 ≥ 16√2/3 .
1Hey , that ' s easy .
Apply Cauchy - Schwarz inequality to a 3 / 2 , b 3 / 2 , c 3 / 2 and a 1 / 2 , b 1 / 2 , c 1 / 2 .
We get , ( a 2 + b 2 + c 2 ) 2 ≤ ( a 3 + b 3 + c 3 ) ( a + b + c )
Now apply C - S to a , b , c and 1 , 1 , 1 .
So , ( a + b + c ) 2 ≤ ( a 2 + b 2 + c 2 ) ( 1 + 1 + 1 )
Hence ,
a 3 + b 3 + c 3 ≥ ( a 2 + b 2 + c 2 ) 2 / ( a + b + c ) = 64√ 3 . 8 = 16 √ 2√ 3
106Inequality not valid unless further conditions given about a,b,c
for eg. a = - √8, b=0, c=0
satisfies a2+b2+c2=8
but doesnt satisfy the inequality required to be proved
39But Cauchy Schwarz proves it that way...strange. I was checking for √8 but you've mentioned a minus sign too. I think it's for positive numbers then?
Nope...Cauchy is for reals. Why then this paradox?
341I guess he forgot the condition that they are all +ve.
Power Mean Inequality gives
\left(\frac{a^3+b^3+c^3}{3} \right)^{\frac{1}{3}} \ge \left(\frac{a^2+b^2+c^2}{3} \right)^{\frac{1}{2}}
from which it follows that
a^3+b^3+c^3 \ge 3 \left( \frac{8}{3} \right)^{\frac{3}{2}} = 16 \sqrt{\frac{2}{3}}
21This is not paradox or anything
when a,b,c are negative a3/2 or a1/2 are not defined.
C.S can only prove |a|3+ |b|3+ |c|3≥16√2/3
So it must be given that terms are positive. Both the proofs are perfect.
