Ask for proof of this inequality, verify this inequality

a^2b+b^2c

a*ab+b*bc)

a(a^2+b^2)/2+b(b^2+c^2)/2(a^3+ab^2+b^3+bc^2)/2...1) In the same way, we get: b 2c + c 2a

b^3+c^3+bc^2+ca^2)/2...2)b^2c+a^2b<=

b^3+a^3+bc^2+ab^2)/2...3) 1) +2) +3) get:

2(a^2b+b^2c+c^2a)<=

a 3+b 3+c 3+a 2b+b 2c+c 2a) So, a 2b+b 2c+c 2a<=a 3+b 3+c 3 i.e., a 3+b 3+c 3>=a 2b+b 2c+c 2a

From the schur inequality, we know a r(a-b)+b r(b-c)+c r(c-a) 0 (r>0) and take r=2 to obtain the original inequality.

Or from the order inequality (order and reverse order sum) just discuss a b c and a c b (because the original inequality is rotationally symmetrical) a b c deduces a 2 b 2 c 2 a c b b deduces a 2 c 2 b 2 and then derives the original inequality from the order and inverse order sum Are the conditions for the problem a, b, c 0? Otherwise, take b=c=0 a<0 then the original inequality does not hold.

The one on the first floor is correct, please see **.

<> used the law of Lopida to prove that the limit of the ratio between the two is 0, that is, the balance of the medium fiber.

Proof : According to the characteristics of the problem, first prove the inequality ln[(n+1)]-ln n>1-n (n+1).

That is, it is proved that ln[(n+1) n]>1-n (n+1) lets the function f(x)=lnx-1+1 x (x>1)f'(x)=1 x-1 x =(x-1) x x>1, f'(x)>0, f(x) is the increment function f(x)> f(1)=0

n+1)/n>1

f[(n+1)/n]>1

i.e. ln[(n+1) n]>1-n (n+1), and then the proved inequality can be obtained by superposition.

x=3-y-z

Substituting the square condition, and then looking at it as a quadratic equation of z, this equation has a real root, and the quadratic discriminant is greater than or equal to 0, and the conclusion 0 can be derived

a²+b²+5-2(2a-b)

a²-4a+4)+(b²+2b+1)

a-2)²+b+1)²

The square is greater than or equal to 0

So (a-2) +b+1) 0

So a + b +5-2 (2a-b) 0

So a + b +5 2 (2a-b).

The left formula subtracts the right formula, and then the recipe can be obtained.

a-2) 2+(b+1) 2 0 is established.

Therefore, the original formula is proven.

Conclusion: The slag is the opposite of the bridge. The gear is chaotic.

This one is very simple, you have to make good use of the formula.

(1) Prove the inequality on the left first:

0 ln(1+x)-[xlnx) (1+x)] x 0) proves that (1+x)ln(1+x)-xlnx 0 receives f(x)=(1+x)ln(1+x)-xlnxf'(x)=ln(1+x)+1-(lnx +1)=ln(1+1/x)

Obviously, when 1, f'(x) 0, f(x) in x (1, + is an increasing function and f(1)=2ln2, so f(x)>f(1)=2ln2>0

i.e. f(x)>0, so (1+x)ln(1+x)-xlnx 0 is thus proved on the left.

2) In the same way, the right side is equivalent to (1+x)lnx-xlnx-2 xln2 0

Let g(x) = (1+x)lnx-xlnx-2 xln2g'(x)=ln(1+1 x)-ln2 xlet g'(x)=0

i.e. ln(1+1 x)-ln2 x

Knowing that x=1 is an extreme point of g(x), when x (0, 1), then g'(x)≥0

When x (1, +, g'(x) 0 x=1 is the maximum point of g(x), and g(1)=0 so g(x) g(1)=0

That is, (1+x)lnx-xlnx-2 xln2 0, so the right side is proved.

In summary, (1) and (2) the original inequality is proven.

Hope it helps!

[[[1]]]

Knowable, x (0, +.)

Let's start by proving the inequality on the left:

0＜ln(1+x)-[xlnx)/(1+x)]∵x＞0.

This inequality can be reduced to:

xlnx＜(1+x)ln(1+x)

When 0 x 1, it is easy to know xlnx 0

and (1+x)ln(1+x) 0

When 0 x 1, there is xlnx(1+x)ln(1+x) When x=1, there is obviously xlnx=0 2ln2=(1+x)ln(1+x) When x 1, the constructor f(x)=xlnx x (1, +

Derivative, f'(x)=(lnx)+1＞0

On (1, +, the function f(x) is incremented, and xlnx (x+1)ln(1+x).

In summary, when x 0, there is always xlnx (x+1)ln(x+1), i.e. 0 ln(1+x)-[xlnx) (x+1)][3]]] thinking.

(a+1/a)²+b+1/b)²

a+1/a+b+1/b)²/2

1+(a+b)/ab)²/2

1+1/ab)²/2

1+1/((a+b)/2)²)/2

25 2 (only if a=b=1 2 is equal sign).

If you see if the title is copied wrongly, when a=b=1 2 is clearly equal to 25 2