x * √( 1+y * y)=√2/2 * √( 2x * x( 1+y * y))& lt； =√2/2*(2x*x+ 1+y*y)/2.....( 1)

You can get x2+y2/2 = 1:2x 2+y2 = 2, so (1) formula < =√2/2*3/2=3√2/4.

2)

Only when x2+3 = 2/ radical sign (x 2+3)) = 2, (x 2+3)+ (under radical sign) ≥ 4 "=" is not true ... (2), and formula (2) is obviously X. =3, so it is wrong;

Sinx+( 1/sinx)=2 is also true if and only if sinx= 1/sinx= 1, but obviously there is no solution at this time, so it is also wrong. In fact, this problem can be solved in other ways, and there are many ways. Here I say a simple one, which is also commonly used by myself, for reference: this kind of problem can be solved by the function fx=x+a/x, (a>0), X.

3)

√( a+ 1/2)+√( b+ 1/2)= √( 1 *(a+ 1/2))+√( 1 *(b+ 1/2))& lt； =( 1+a+ 1/2)/2+( 1+b+ 1/2)/2 =(3+a+b)/2 = 2；

√(a+ 1/2)+√(b+ 1/2)>= √( 2 √( 1/2a))+√( 2 √( 1/2b))....(3)

(3) the formula "=" holds if and only if a=b= 1/2, so (3) >; = √ (2 * √ (1/2 *1/2)+√ (2 * √ (1/2 *1/2) =1,that is1

4) I am calculating, and I feel that you may have typed the question wrong. It should be cos2x instead of cosx.

5)

1/X+ 1/Y = 1/30((4X+Y)/X+(4X+Y)/Y)

= 1/30 *(5+Y/X+4X/Y)≥ 1/30 *(5+2√4)= 3/ 10

The equal sign holds when Y/X=4X/Y, that is, Y=2X, and it is obtained by substituting 4X+Y=30.

X=5，Y= 10

6)

Note that the range of values is r, which means that fx can get all positive numbers.

So let gx=x+a/x-4, then the minimum value of gx should be less than or equal to 0, because A >；; 0,

x+a/x & gt； = 2√a & gt； 4, so gxmin = 2 √ A-4.

Get a < = 4,

To sum up, 0 < a<1∪1< a & lt=4