It is known that a and B are real numbers, and e < a < B, where e is the base of natural logarithm

It is known that a and B are real numbers, and e < a < B, where e is the base of natural logarithm

Proof: when e < a < B, to prove AB > Ba, as long as blna > AlNb is proved, that is, as long as LNAA > lnbb is proved, consider the function y = lnxx (0 < x < + ∞) because when x > e, y ′ = 1 − lnxx2 < 0, so the function y = lnxx is a decreasing function in (E, + ∞), because e < a < B, so LNAA > lnbb, that is ab > ba

How to prove n ^ N.E ^ m > = m ^ N.E ^ n? (E is the base of natural logarithm)

It is proved that when x > 0, there is e ^ x > = ex
When x = 1, the equal sign holds,
It can be seen from the derivation
x> At 1, e ^ x increases faster than ex
When x changes from 1 to 0, e ^ x decreases more slowly than ex,
So the above inequality holds
Let X be m / n,
There is e ^ (M / N) > = (M / N) e,
ne^(m/n)>=me
The objective inequality is obtained to the nth power of both sides at the same time
If you don't understand, please ask

Given a + B = 6, ab = 8, find the value of a square + b square, (a-b) square and a square - b square respectively

a²+b²=(a+b)²-2ab=36-16=20
(a-b)²=(a+b)²-4ab=36-32=4
A-B = 2 or - 2
The value of a-square-b-square
=(a+b)(a-b)
=12 or - 12

a. B, C belong to R, a > 0, b > 0, 2C > A + B, C square and ab size relationship

2c>a+b ,a>0,b>0
4c^2>(a+b)^2=a^2+b^2+2ab>=2ab+ab=4ab
c^2>ab

Given that A-B = 4, ab = 3, find the square of a + B and the square of (a + b) and ask you to 3Q

(a-b)^2=4^2=16 ab=3 (a-b)^2=a^2-2ab+b^2=16 a^2+b^2=16+2x3=22 (a+b)^2=a^2+2ab+b^2=22+6=28

Give three integral A's Square, B's square sum 2Ab

(1) When a = 3, B = 4, A2 + B2 + 2Ab = (a + b) 2 = 49
(2) If A2, B2 are chosen, then A2-B2 = (a + b) (a-b)
If a 2,2 AB is selected, then a 2 ± 2 ab = a (a ± 2 b)

The common factor of the quadratic power of a-2ab + B and the quadratic power of A-B

(a-b)^2=a^2-2ab+b^2 ; a^2-b^2=(a-b)(a+b)

Given the square of a-Ab = 21ab-b = - 15, find the square of A-B and the square of a-2ab + B

a²-b²=(a²-ab)+(ab-b²)=21-15=6
a²-2ab+b²=(a²-ab)-(ab-b²)=21-(-15)=21+15=36

Given that the square of a + AB = 3, the square of AB + B = 7, find the value of (1) the square of a = the square of 2Ab + B, and (2) the square of A

a^2+ab=3,ab+b^2=7
a^2+ab+ab+b^2=7+3=10=a^2+2ab+b^2=(a+b)^2
(a + b) ^ 2 = 10, then a + B = √ 10
a^2=2ab+b^2,a^2=√10+b^2+ab=√10 +7

A + B + C + D = 20 a square + 4B square + 9C square + 16d square = 120 a.b.c.d is a positive real number

A + B + C + D = 20A ^ 2 + 4B ^ 2 + 9C ^ 2 + 16d ^ 2 = 120 a.b.c.d belongs to positive real number. Try to find the maximum value of d by using Cauchy inequality (a ^ 2 + 4B ^ 2 + 9C ^ 2) [1 ^ 2 + (1 / 2) ^ 2 + (1 / 3) ^ 2] ≥ ((a + B + C) ^ 2A + B + C = 20-da ^ 2 + 4B ^ 2 + 9C ^ 2 = 120-16d ^ 2), so (120-16d ^ 2) * 49 / 36 ≥ (20-d) ^ 2 come on, wish