341calculus sort of kills the fun. any algebraic method?
8^{\frac{1}{n}} + 27^{\frac{1}{n}} + 64^{\frac{1}{n}} > 12^{\frac{1}{n}} + 32^{\frac{1}{n}} + 36^{\frac{1}{n}}
for n>1
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10 Answers
9we have to prove
f(x)=8x+27x+64x-12x-32x-36x > 0
for x≡(0,1)
f(0)=0 and f(1) >0
f'(x) is -ve for x≡(0,1) and f'(x)=0 for x=0
1357same as for n<1
8^{n}+27^{n}+64^{n} > 12^{n}+32^{n}+36^{n}
a=2
b=3
c=4
Thus the clue is
a^3+b^3+c^3>2(ab+bc+ca)
but i cant unjumble the clue :(
1let\ 2^{\frac{1}{n}}=a,3^{\frac{1}{n}}=b,4^{\frac{1}{n}}=c\\ for\ every\ n>1,we\ have\ c>b>a\\ so\ by\ rearrangement\ inequality\ we\ have,\\ a^3+b^3+c^3 > a^2b+b^2c+c^2a \\ so\ we\ have\\ 8^{\frac{1}{n}}+27^{\frac{1}{n}}+64^{\frac{1}{n}}>12^{\frac{1}{n}}+32^{\frac{1}{n}}+36^{\frac{1}{n}}
do correct if something is wrong
9wrt #4 i cudnt get u sir a=2 ???
algebraic method ( n≡(0,1) )
A= (3/2)n
B= 2n
eq reduces to
1+A3+B3-A-A2B-B2 > 0
ie (B2-A2)(B-1) + (A2-1)(A-1) > 0
which is true always as B>A>1
9@ dimensions sry i dint see ur post b4
i proceeded same way as uve done and then divided whole expression by a3
and substituted A=b/a
B=c/a
and then derived manually
i guess the abv is proof of the rearrangement inequality that u r talking about
9@ prophet sir
its not always easy unless we know that there exists something called a rearrangement inequality
9ok
sorry
dint know ppl are supposed to know these in 12th itself
pls comment upon the right ans after someone types in this post and all ur other posts sir