Big y photography series junior high school students' seeds

I don't know your age, so I'm not sure how to solve this problem. Please forgive me.

1) If you are a junior high school student, you can answer with the knowledge of quadratic function.

Solution: let x+2y+2z=k, then x=k-2y-2z is substituted into x? +y？ +z？ = 1

Get (k-2y-2z)？ +y？ +z？ = 1

(k？ +4y？ +4z？ -4ky-4kz+8yz)+y？ +z？ = 1

Take y as the principal component

Get 5y？ -4ky+8zy+5z？ -4kz+k？ - 1=0

That is, 5y? -(4k-8z)y+5z？ -4kz+k？ - 1=0

Because y is a real number, that is, this equation has a real number root.

△=(4k-8z)？ -4×5×(5z？ -4kz+k？ - 1)≥0

4(k-2z)？ -5(5z？ -4kz+k？ - 1)≥0

4k？ - 16kz+ 16z？ -25z？ +20kz-5k？ +5≥0

-9z？ +4kz-k？ +5≥0

Let f(z)= -9z? +4kz-k？ +5

Then the vertex of parabola f(z) with downward opening must be above or on the X axis.

That is, the ordinate of the vertex is greater than or equal to 0.

That is 4×(-9)×(-k? +5)-(4k)？ /[4×(-9)]≥0

Quadratic function y=ax? +bx+c (the vertex of-b/2a, [4ac-b? ]/4a)

20k？ - 180≤0

k？ ≤9

-3≤k≤3

So the maximum value of x+2y+2z is 3.

2) If you are a high school student, it is more convenient to use Cauchy inequality.

Three-dimensional Cauchy inequality: (a? +b？ +c？ )(x？ +y？ +z？ )≤(ax+by+cz)？ (and only take an equal sign when a/x=b/y=c/z)

So (1? +2? +2? )(x？ +y？ +z？ )≤( 1×x+2×y+2×z)？

That is, 9× 1≤(x+2y+2z)?

-3≤x+2y+2z≤3

So the maximum value of x+2y+2z is 3.

Given the sum of squares of several quantities (the coefficient is not 1), the range of any linear combination of these quantities can be found by Cauchy inequality.

Math lovers will answer your questions wholeheartedly.