When x > 0, the inequality X / E + X is proved

When x > 0, the inequality X / E + X is proved

Lnex=1+lnx
First prove that LNX makes f (x) = x-lnx-1, (x > 0)
As long as it is proved that the minimum value of F (x) is greater than zero, it is proved that X-1 > LNX
F'(x)=1-1/x,F'(x)>0==>x>1,F'(x)<0==>0x=1
That is, f (x) is a subtractive function on (0,1) and an increasing function on (1, + 8). When x = 1, the minimum value f (1) = 1-ln1-1 = 0. When x > 0, f (x) > = 0 is always true!
In fact, when you draw the image of y = LNX, you will find that its tangent at point (1,0) is y = X-1, and the whole image of y = LNX is below the image of y = X-1, that is, X-1 > = LNX is always true!
x/e+x[(1/e)+1]x-1?
It's impossible! Let x = e ^ 2 and you can overthrow it
About this old problem, you usually need to prove: 1-1 / X=

Prove the inequality: when x > 0, e ^ x > 1 + X + x ^ 2 / 2 1. Prove inequality: when x > 0, e x > 1 + X + x 2 / 2

It is proved that if f (x) = e ^ X - (1 + X + x ^ 2 / 2), then f '(x) = e ^ X - (x + 1) f' '(x) = e ^ X-1, it is easy to know that f' '(x) monotonically increases on R. therefore, when x > 0, f' '(x) > F' '(0) = 0, then f' '(x) monotonically increases on (0, + ∞); then f' '(x) > F' '(0) = 0, and it is deduced that f (x) also increases on (0, + ∞)

Let the function f (x) = (A / 2) x ^ 2 (a ≠ 0), G (x) = x + 1 / e ^ X. it is proved that when a ≥ 1, the inequality (1 - (A / 2) x ^ 2) e ^ x ≤ x + 1 For any x belonging to [0, + ∞) is constant

Let y = f (x) + G (x)
That is, y ≥ 1 is constant for any x ∈ R (a ≥ 1)
y'=x+2/e^x +ax
y''=x+3/e^x +a＞0
So y 'monotonically increases on R
When x tends to negative infinity, y '< 0
When x tends to positive infinity, y '> 0
So y '= 0 has and has only one solution, which is written as X
That is X. + 2 / e ^ X. + ax. = 0 [when x = 0, y '= 2 ＞ 0, so X. ＜ 0]
So y decreases monotonically in (- ∞, X.) and increases monotonically in (X., + ∞)
So y ≥ X. + 1 / e ^ X. + (A / 2) X. ^ 2
=x.+2/e^x.+ax.-ax.-1/e^x.+（a/2）x.^2
=-ax.-1/e^x.+（a/2）x.^2
Note m (X.) = - ax. - 1 / e ^ X. + (A / 2) X. ^ 2
m'=ax.+1/e^x.-a＜0
So m single adjustment and reduction
So m (X.) > m (0) = y (0) = 1
So y ≥ 1
So the original proposition is proved
The logical meaning of greater than or equal is greater than or equal to. Here '' = '' can not be taken actually, but the proposition is correct
PS: if you don't understand, ask again. The college entrance examination has ended recently. There is a lot of time
It's a little strange to finish this problem. It seems that a > 0 can be done. Maybe I'm wrong? Or which bracket? What didn't you write?

It is proved that when x > 0, e ^ x > x + 1

Note f (x) = e ^ x-x-1
Then f (0) = 0
When x > 0, f '(x) = e ^ X-1 > 0
Therefore, f (x) is an increasing function at x > O, so f (x) > F (0) = 0, that is, e ^ x > x + 1

Prove the inequality e ^ x > ex (x > 1). Thank you very much! It's best to use extreme value.

Let f (x) = e ^ x-ex (x > 1)
f'(x)=e^x-e
When x > 1, there are: e ^ x > e, that is: F '(x) > 0
So f (x) is monotonically increasing in x > 1
Because: F (1) = E-E = 0
So:
When x > 1: F (x) > F (1) = 0
E ^ x > ex
The original question is proved

What is the height of the largest cone inscribed on a sphere with radius r? What is the height of the largest cone inscribed on a sphere with radius r?

4R / 3: if the height of the cone connected to the ball is h, the radius of the cone bottom is p, P ^ 2 = R ^ 2 - (H-R) ^ 2 = 2rh-h ^ 2, and the volume is v = 3.14 * (2rh-h ^ 2) * H / 3. Take its derivative. When v ` = 0, i.e. H = 4R / 3, the volume takes the extreme value (also the maximum value); At this time, v = 3.14 * [2R * 4R / 3 - (4R / 3) ^ 2] * (4R / 3) / 3 = 3.14 * 8 * R ^ 3 / 27 = 2 * V ball / 9