how demonstrate integral (1_2)e^x 1/x dx <e^4-e^2+1)/2?

with 2x y <x^2+y^2

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You need to prove the given inequality, using the `2xy <= x^2 + y^2` , hence, you may consider two functions, such that:

`f(x) = e^x, g(x) = 1/x`

Evaluating the following square yields:

`(f(x) - g(x))^2 >= 0`

`f^2(x) - 2f(x)*g(x) + g^2(x) >= 0`

`f^2(x) + g^2(x) >=2f(x)*g(x) => (f^2(x) + g^2(x))/2 >= f(x)*g(x) = e^x*(1/x)`

Integrating the product `f(x)*g(x)` , over the interval `[1,2]` , yields:

`int_1^2 e^x*(1/x) dx = int_1^2 f(x)*g(x) dx <= (1/2)int_1^2 (f^2(x) + g^2(x)) dx`

Using the property of lineariy of integral, yields:

`int_1^2 e^x*(1/x) dx<= (1/2)(int_1^2 f^2(x) dx) +(1/2)( int_1^2 ` `g^2(x) dx)`

`int_1^2 e^x*(1/x) dx<= (1/2)(int_1^2 e^(2x) dx + int_1^2 1/(x^2) dx)`

`int_1^2 e^x*(1/x) dx<= (1/2)(e^(2x)/2|_1^2 - 1/x|_1^2)`

Using the fundamental theorem of calculus yields:

`int_1^2 e^x*(1/x) dx<= (1/2)(e^4/2 - e^2/2 - 1/2 + 1)`

`int_1^2 e^x*(1/x) dx<= (1/2)(e^4/2 - e^2/2 +1)`

`int_1^2 e^x*(1/x) dx<= (e^4 - e^2 + 1)/4`

**Hence, testing the given inequality, using `f^2(x) + g^2(x) >=2f(x)*g(x)` , yields that `int_1^2 e^x*(1/x) dx<= (e^4 - e^2 + 1)/4` holds for f(x) = e^x and **`g(x) = 1/x.`

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