inequalities..

Hey , 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

Inequality 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

I 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}}

yeah the furthur condition to be imposed is a ,b,c>=0

This is not paradox or anything

when a,b,c are negative a^3/2 or a^1/2 are not defined.

C.S can only prove |a|³+ |b|³+ |c|³â‰¥16√2/3

So it must be given that terms are positive. Both the proofs are perfect.