IIT JEE , AIEEE Forum - Mathematics , Algebra - prove that 5 > sum of square root 5,cube root 5 &fourth root 5

ds

prove that 5 > sum of square root 5,cube root 5 &fourth root 5

challenger

I am also stuck up with olympiad question.

challenger

question is wrong .
It is 5

(**328**)

nadeemoidu

³

5 + ⁴

5

How can you say that (⁴

5 +

5 )/2 > ³

5 ?

It cannot be done by AM>GM.

Also 3 ³

5 =/= ³

125 .

I think the answer given by 12121212 is completely wrong.

goutam.chalasani

wat nadeemoidu said is right u have to prove ur last step which u took for granted

spideyunlimited

LMFAO DUDE... ur question is wrong. the question was to prove that 5 is LESS than root 5 + cube root 5 + quadroot 5.

rshm15varma

hehehe...yup question is wrong.

me

see people this can be done by taking x=f(x) & x^1/2+x^1/3+x^1/4=g(x) and proving g>f for x=5.

and the shorter method is -------
5^1/2 > 4^1/2=2
5^1/3 > (125/27)^1/3=5/3 > 1.6
5^1/4 > 4^1/4=2^1/2 > 1.4
hence 5^1/2 + 5^1/3 + 5^1/4 >2+1.4+1.6 = 5

rate if u like it

nadeemoidu

Complete RMO solutions http://www.isid.ac.in/~rbb/crmosol_07.pdf

priyesh

here's the simplest solution to prove

5 < sum of square root 5,cube root 5 &fourth root 5

using just basic arithmetic

see

sqrt(5) > 2.2 (as (2.2)^2 = 4.84)

cuberoot(5) > 1.6 (as (1.6)^3 = 4.096)

fourthroot(5) = root(root(5)) > sqrt(2.2) >sqrt(2) => fourthroot(5) > 1.4

therefore

[sum of square root 5,cube root 5 &fourth root 5] > 2.2 + 1.6 + 1.4 = 5.2

since it is greater than 5.2 hence it is greater than 5

Hope you all like this method

cheers!!!!!!!!!!!

johri_anshuman

Heres another method that proves all the inequalities in that question

let f(x) =

x + ³

x + ⁴

x

and g(x) = x

now just check the values of f(x) for different values of x. we know that f(x) is strictly increasing function. plot the curve.

f(x) and g(x) intersect for some x approx = 5.7

now just graphically check the inequalities.

g(5)<f(5) proves first inequality,

g(8)>f(8) proves second inequality,

g(x)>f(x) for all x>9 proves the third inequality.

nadeemoidu

We were not provided with graph plotters for the regional mathematics olympiad :)

johri_anshuman

hey i plotted that graph just by checking the values in dat exam. the only problem was that i could not get any solution to f(x)= g(x) and still i managed to get the solution around 6. u dont need any graph plotters.

spideyunlimited

priyesh ur answer is the one that is given by the olympiad officials on the link provided by the person above ur post.

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