Do not use calculus!

Algebra Level 4

Let

A = ( x + 1 x ) 6 − ( x 6 + 1 x 6 ) − 2 ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) A = {\frac{(x + \frac{1}{x})^{6} - (x^{6} + \frac{1}{x^{6}}) - 2}{(x + \frac{1}{x})^{3} + (x^{3} + \frac{1}{x^{3}})}}

Find m i n ( A ) min(A) .

Clarifications: x x is a positive real number.


The answer is 6.

Sanchayapol Lewgasamsarn by Sanchayapol Lewgasamsarn , 20, Thailand

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4 solutions

Let x + 1 x = n x + \frac{1}{x} = n , then x 3 + 1 x 3 = n 3 − 3 n x^{3} + \frac{1}{x^{3}} = n^{3} - 3n

Then, x 6 + 1 x 6 = ( n 3 − 3 n ) 2 − 2 x^{6} + \frac{1}{x^{6}} = (n^{3} - 3n)^{2} - 2 .

So, by subsituting the values of n n into A A , the result will be 3 n \boxed{3n} .

The value of A A is 3 ( x + 1 x ) 3(x + \frac{1}{x}) .

And, by the A.M. - G.M. inequality,

3 ( x + 1 x ) ≥ 6 ⋅ x ⋅ 1 x 3(x + \frac{1}{x}) \geq 6 \cdot \sqrt{x \cdot \frac{1}{x}}

3 ( x + 1 x ) ≥ 6 3(x + \frac{1}{x}) \geq 6

The minimum value of A A is 6 \boxed{6}

The minimum value will be reached when x = 1 x x = \frac{1}{x} .

11 Helpful 0 Interesting 0 Brilliant 0 Confused

Please add in the conditions that x x is a positive real number.

Jared Low - 7 years, 1 month ago

Where is am inequality u HV written same value 3n only in terms of x .. And how can u say that am is more than gm

Rohit Singh - 7 years, 1 month ago

[ ( x 2 + 1 ) 6 − ( x 6 + 1 ) 2 ] [ ( x 2 + 1 ) 3 + ( x 6 + 1 ) ] ∗ x 3 \frac{[(x^{2}+1)^{6}-(x^{6}+1)^{2}]}{[(x^{2}+1)^{3}+(x^{6}+1)]*x^{3}} = [ ( x 2 + 1 ) 3 − ( x 6 + 1 ) ] ∗ [ ( x 2 + 1 ) 3 + ( x 6 + 1 ) ] [ ( x 2 + 1 ) 3 + ( x 6 + 1 ) ] ∗ x 3 =\frac{[(x^{2}+1)^{3}-(x^{6}+1)]*[(x^{2}+1)^{3}+(x^{6}+1)]}{[(x^{2}+1)^{3}+(x^{6}+1)]*x^{3}} = [ ( x 2 + 1 ) 3 − ( x 6 + 1 ) ] x 3 =\frac{[(x^{2}+1)^{3}-(x^{6}+1)]}{x^{3}} = [ x 6 + 3 x 4 + 3 x 2 + 1 − ( x 6 + 1 ) ] x 3 =\frac{[x^{6}+3x^{4}+3x^{2}+1-(x^{6}+1)]}{x^{3}} = ( 3 x 4 + 3 x 2 ) x 3 =\frac{(3x^{4}+3x^{2})}{x^{3}} = 3 [ x + ( 1 / x ) ] > = 3 ∗ 2 = 6 ⇒ M i n ( A ) = 6 =3[x+(1/x)]>=3*2=6 \Rightarrow Min(A)=\boxed{6}

aaaaaa bbbbbb - 7 years, 1 month ago

When x =1 ,A=6

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Phudis Dawieang
Apr 20, 2014

We know the truth that x + 1 x ≥ 2 x+\frac { 1 }{ x } \ge 2 (proof by AM-GM) So,you just substitute x + 1 x = 2 x+\frac { 1 }{ x } = 2 and x 6 + 1 x 6 = 2 { x }^{ 6 }+\frac { 1 }{ { x }^{ 6 } } =2 and you'll get 2 6 − 2 − 2 2 3 + 2 = 6 \frac { { 2 }^{ 6 }-2-2 }{ { 2 }^{ 3 }+2 } \quad =6 #

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Rajen Kapur
Apr 16, 2014

The question should mention that x is positive real.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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