The convergence of three Cauchy inequalities and mean inequalities 1. Solve the inequality system: y ^ 2-2ax0, note a > 0 (just write down the main process, the result may not be very good) 2 if there is: (x / y ^ 2 + Z ^ 2) + (Y / x ^ 2 + Z ^ 2) + (Z / y ^ 2 + x ^ 2) > = K / radical (x ^ 2 + y ^ 2 + Z ^ 2) Find the maximum value of K (the maximum is (root 27) / 2, how come?) 33 verification: x ^ 2 / y ^ 2 + Z ^ 2 + YZ) + (y ^ 2 / x ^ 2 + Z ^ 2 + XZ) + (Z ^ 2 / y ^ 2 + x ^ 2 + XY) > = 1 (no idea...)

The convergence of three Cauchy inequalities and mean inequalities 1. Solve the inequality system: y ^ 2-2ax0, note a > 0 (just write down the main process, the result may not be very good) 2 if there is: (x / y ^ 2 + Z ^ 2) + (Y / x ^ 2 + Z ^ 2) + (Z / y ^ 2 + x ^ 2) > = K / radical (x ^ 2 + y ^ 2 + Z ^ 2) Find the maximum value of K (the maximum is (root 27) / 2, how come?) 33 verification: x ^ 2 / y ^ 2 + Z ^ 2 + YZ) + (y ^ 2 / x ^ 2 + Z ^ 2 + XZ) + (Z ^ 2 / y ^ 2 + x ^ 2 + XY) > = 1 (no idea...)

1. Combination of numbers and shapes. Y ^ 2-2ax0, outside the circle. Notice that the circle and the parabola are "tangent" (in fact, only one intersection point is enough). So the answer is: the parabola goes out of the circle. (note that the lower boundary should not be wrong) 2. Consider the most

This inequality is proved by Cauchy inequality
It is known that Xi ≥ 0 (I = 1,2,3,...) ,n),√(x1+x2+…… +xn)(x1^3+x2^3+…… +xn^3)≥x1^2+x^2+…… +xn^2

xi=(xi^0.5)^2
xi^3=(xi^1.5)^2
The original formula can be changed into
【Σ(xi^0.5)^2 】【Σ(xi^1.5)^2】
>=【Σ(xi^0.5)(xi^1.5) 】^2
=【Σ（xi^2)】^2
Both sides can be opened at the same time

First aid ~ SOS
If x > A + 2 has no solution, try to judge the solution of x > 2 - A
x

x>a+2
x2-a>0
x

Trigonometric inequality
Set 0

Trigonometric inequality
Set 0