In the application of higher one inequality, a > 0, b > 0, it is proved that 2 (radical a + radical b) ≤ a + B + 2

Because (√ A-1) ^ 2 + (√ B-1) ^ 2 > = 0
So A-2 √ a + 1 + B-2 √ B + 1 > = 0
a+b+2>=2√a+2√b
SO 2 (√ a + √ b)

The following inequality a / radical B + radical B ≥ 2 radical a is proved The following inequality a / radical (b) + radical B ≥ 2 radical a (a, B ∈ R +) is proved

Obviously, x 2 + y 2 ≥ 2XY
Equal sign only if x = y
Because both a and B are greater than 0, we can make a / √ B = x 2 √ B = y 2
x²+y²=(a／√b)+√b≥2√[(a／√b)·√b]=2√a
Only if a / √ B = √ B, i.e. a = B, take the equal sign
In addition, the mean inequality m + n ≥ 2 √ Mn can also be used directly

High school inequality proof (a ^ 2 + AB + B ^ 2) ^ 1 / 2 + (b ^ 2 + AB + C ^ 2) ^ 1 / 2 > = a + B + C

If a, B, C > 0, and Jensen inequality f (x1 + x2 +. Xn) > = f (x1) +. + F (xn) takes function y = x ^ 1 / 2 (1 / 2) (x > 0), we can get (a ^ 2 + AB + B ^ 2) ^ 1 ^ 2 = f (a ^ 2 + AB + B ^ 2) > F (a ^ 2 + AB + B ^ 2) > F (a ^ 2) + F (AB) + F (AB) + F (b ^ 2) > A + B + root (AB) (b ^ 2 + AB + C ^ 2) ^ (1 / 2) = f (b ^ 2 + AB + C ^ 2) > F (b ^ 2 + B ^ 2) > F (b ^ 2) + F (b ^ 2) + F (b ^ 2) + F (b ^ 2) + F (f (AB) +

Prove the inequality 2 / (1 / A + 1 / b) ≤√ AB by analytic method

Obviously, the condition of the title is insufficient
If you add "a > 0, b > 0", then
2/(1/a+1/b)≤√ab
A kind of 2ab/(a+b)≤√ab
A kind of 2(√ab)²≤(√ab)(a+b)
A kind of 2√ab≤a+b
A kind of (√a-√b)²≥0.
The above formula holds and every step is reversible
So the original inequality holds

How to prove that ABC is less than or equal to one 27th of the third power of (a + B + C) ABC ≤ 1 / 27 (a + B + C) 3

A, B, C are positive real numbers. A ^ 3 + B ^ 3 + C ^ 3-3abc = (a + B + C) (a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ac) = 0.5 (a + B + C) (2a ^ 2 + 2B ^ 2 + 2C ^ 2-2ab-2bc-2ac) = 0.5 (a + B + C) [(a-b) ^ 2 + (B-C) ^ 2 + (B-C) ^ 2 + (C-A) ^ 2] a + B + C > 0, (a-b) ^ 2 + (B-C) ^ 2 + (C-A) ^ 2] a + B + C > 0, (a-b) ^ 2 + (a-b) ^ 2 + (a-b) ^ 2 + (a-b) ^ 2 + (a-b) ^ 2 B-C) ^ 2 + (C-A) ^ 2 ≥ 0

It is proved that inequality 1 / A + 1 / B + 1 / C ≥ 9 / π in triangle ABC

prove:
∵ π / 3 = (a + B + C) / 3 ≥ cubic root ABC
ν 1 / cubic root ABC ≥ 3 / π
ν 1 / A + 1 / B + 1 / C ≥ 3 / cubic root ABC ≥ 9 / π

High school inequality 1 / (a ^ 3) + 1 / (b ^ 3) + 1 / (C ^ 3) + ABC is greater than or equal to 2 √ 3

1 / (a ^ 3) + 1 / (b ^ 3) + 1 / (C ^ 3) + 1 / (C ^ 3) ≥ 3 {[1 / (a ^ 3)] * [1 / (b ^ 3)] * [1 / (b ^ 3)] * [1 / (C ^ 3)]} ^ (1 / 3) = 3 [1 / (a ^ 3 * B ^ 3 * C ^ 3)] ^ (1 / 3) = 3 / (ABC), so 1 / (a ^ 3) + 1 / (b ^ 3) + 1 / (C ^ 3) + ABC ≥ 3 / (ABC) + (ABC) ≥ 2 {[3 / (ABC)] * (ABC)} (1 / 2) = 2 * 3 ^ 3 ^ (1 / 2) = 2 * 3 ^ (1 / 2) = 2 * 3 ^ 1 * 3 ^ 1 (1 / 2) = 2 * 3 ^ 1 / (ABC)] * (ABC)} (ABC)}/ 2) = 2 √ 3

It is proved that ABC ^ 3 is less than or equal to 27 ((a + B + C) / 5) ^ 5 for any positive real number a, B, C In fact, this is the second question, I will give the original question, I hope some help. The original question: F (x, y, z) = LNX + LNY + lnz on the sphere x ^ 2 + y ^ 2 + Z ^ 2 = 5 ^ 2 (where x, y, Z are all positive numbers). This is the first question. I finished it. Next, it asks me to prove the second question I asked

Let a = x ^ 2 b = y ^ 2 C = Z ^ 2

Given the triangles a, B, C, it is proved that ABC > = (a + B-C) (a + C-B) (B + C-A)

It is proved that ∵ a, B, C are the three sides of ∵ ABC
∴a+b-c>0
a+c-b>0
b+c-a>0
∵(a+b-c)(a+c-b)=a^2-(b-c)^2≤a^2
∴(a+b-c)(a+c-b)(b+c-a)≤a^2(b+c-a)………… （1）
∵(a+b-c)(a+c-b)≤b^2
∴(a+b-c)(a+c-b)(b+c-a)≤b^2(a+c-b)………… （2）
∵(a+b-c)(a+c-b)≤c^2
∴(a+b-c)(a+c-b)(b+c-a)≤c^2(a+b-c)………… （3）
∵ (1), (2), (3) both sides of the three equations are  0
(1), (2), (3) the two sides of the three formulas are multiplied respectively to obtain:
[(a+b-c)(a+c-b)(b+c-a)]^3≤a^2b^2c^2(a+b-c)(a+c-b)(b+c-a)
That is: ABC ≥ (a + B-C) (a + C-B) (B + C-A)

In the triangle ABC, prove a ^ 2 (B + C-A) + B ^ 2 (c + a-b) + C ^ 2 (a + B-C)

Solution 1: sorting inequality
Let a ≥ B ≥ C
It is known that a (B + C-A) ≤ B (c + a-b) ≤ C (a + B-C),
Ordering inequality: reverse order is less than disorder order
a2(b+c-a)+b2(c+a-b)+c2(a+b-c)≤ba(b+c-a)+cb(c+a-b)+ac(a+b-c)
a2(b+c-a)+b2(c+a-b)+c2(a+b-c)≤ca(b+c-a)+ab(c+a-b)+bc(a+b-c)
Adding two formulas
2[a2(b+c-a)+b2(c+a-b)+c2(a+b-c)]≤ba(b+c-a)+cb(c+a-b)+ac(a+b-c)
+ca(b+c-a)+ab(c+a-b)+bc(a+b-c)
=b^2a+abc-a^2b+c^2b+abc-b^2c+a^2c+abc-c^2a+abc+c^2a-a^2c+abc+a^2b-ab^2+abc+b^2c-bc^2=6abc
So a ^ 2 (B + C-A) + B ^ 2 (c + a-b) + C ^ 2 (a + B-C)

In the application of higher one inequality, a > 0, b > 0, it is proved that 2 (radical a + radical b) ≤ a + B + 2