A 0, B 0, determine the size of B2 A and 2B A

To determine the size of these two numbers, just subtract the last one from the previous one.

b2 a-(2b-a)=b2 a-2b+a (b2-2ab+a2) a = (b-a)2 a because a 0 (b-a) 0 so b2 a 2b-a if and only if a=b go to the equal sign hope.

Solution: because: b 0 5 a-(2b-a)=(b 0 5-2ab+a 0 5) a=(b-a) 0 5 a

And because, a 0 , b 0

So: (b-a) 0 5 a 0

i.e., (b 0 5 a) (2b-a).

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

Because a>0, b>0, (1) when a is not equal to b, (b-a) 2 a> 0, so b 2 a > 2b-a

2) When a is equal to b, (b-a) 2 a=0, so b 2 a = 2b-a

Take the logarithm of the comparison value of the lead: a a*b b=a a+b b (ab) [a+b) 2]=(a+b) 2( a+ b) subtract the two logarithms and compare (a a+b b)-=a-b) 2 a-(a-b) 2 b=(a-b) 2 * a b If a>b, a-b>0 a b>1 know a b>0 Formula >> 0....

Because A 2 Rock Chaos B B 2 A> = 2 root number (ab), A B > = 2 root number (ab), and A is not equal to B, so the coarse file A 2 staring B B 2 A > A B

[Method 1] (a 2 b+b 2 a)-(a+b)=[(b-a) 2*(b+a)] (ab)>0

So a b + b a > a + b.

Method 2] a>0, b>0, a≠b, (a 2 b+b) > 2 under the root number (a 2 b*b) = 2a, b 2 a+a) > 2 under the root number (b 2 a*a) = 2b, the two formulas are added: a 2 b + b 2 a + a> 2a + 2b, so a b + b a > a + b.

Solution: From the problem and the basic inequality, it can be known:

a²/b)+b＞2a.

b²/a)+a＞2b.(a≠b), these two inequalities cannot be equated. ）

a²/b)+(b²/a)＞a+b

A 2 B+B 2 A-(A+B)=(A+B)(A-B) 2 Ab Because "0,B>0, and A≠B, So) (A+B)(A-B) 2 Ab>0So, A 2 B+B 2 A-(A+B)>0, So, A 2 B+B 2 A>A+B

(a^2/b+b^2/a)-(a+b)

a³+b³-a²b-ab²)/ab

a²(a-b)+b²(b-a)]/ab=(a-b)(a²-b²)/ab

a-b)(a-b)(a+b)/ab

a-b)²(a+b)/ab

In the above equation, all 3 terms are greater than 0, so (a2 b+b 2 a) (a+b).

by a 0, b 0, |a|＜|b|

It can be seen that A is positive, B is negative, and A is closer to the origin.

a+b＜0，b-a＜0，a+b|=-a-b，b-a|=a-b，（-a-b）-（a-b）

a-b-a+b

2a 0, i.e. |a+b|＜|b-a|。

Two methods:

1) Do the bad method:

ab²-a²b=ab（b-a）

A 0, b 0, ab 0, then.

b a, b-a 0, ab -a b 0, ab a b b=a, b-a=0, ab -a b=0, ab = a b b a, b-a 0, ab -a b 0, ab a b (2) do business law.

ab²/a²b=b/a

b＞a，b/a＞1，ab²＞a²b

a＜b，b/a＜1，ab²＜a²b

a=b，b/a=1，ab²=a²b

Categorical discussion is sufficient.

a²b-ab²=ab(a-b)

a>0, b>0 are known

When a>b ab(a-b)>0 i.e. a b>ab when a=b, a b=ab

When a< b a b

∵a>b>0

a²＞ab＞0

i.e.: a -ab 0 and ab 0

a² +1/ab + 1/a(a-b)

a +1 ab + 1 (a -ab) -ab+ab=[(a -ab)+1 (a -ab)] ab+1 (ab)] 2+2 [basic inequality].

4 If and only if a -ab = 1, ab = 1 take the equal sign, that is: when a = 2, b = 1 2, the original formula has a minimum value of 4