What is the minimum value of X + y? Let x ^ 2 + y ^ 2 = 4 find the minimum value of XY + 4 (x + y) - 2~

What is the minimum value of X + y? Let x ^ 2 + y ^ 2 = 4 find the minimum value of XY + 4 (x + y) - 2~

1. Let t = x + y
Because x, Y > 0,
From the basic inequality,
(x +y) /2 ≥√xy,
That is, XY ≤ (T ^ 2) / 4,
If and only if x = y = t / 2
SO 2 = x + y + XY
≤t +(t^2) /4,
That is T ^ 2 + 4T - 8 ≥ 0
The solution is t ≤ - 2 - 2 √ 3 or t ≥ 2 √ 3 - 2
So x + y ≥ 2 √ 3 - 2,
If and only if x = y = √ 3 - 1, the equal sign holds
That is, when x = y = √ 3 - 1, x + y has a minimum value of 2 √ 3 - 2
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Substitution method, basic inequality, solution inequality
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2. From the mean inequality,
√[ (x^2 +y^2) /2 ] ≥ (x +y) /2 ≥ √xy,
If and only if x = y, the equal sign holds
And because x ^ 2 + y ^ 2 = 4,
So x + y ≤ 2 √ (4 / 2) = 2 √ 2,
xy ≤4/2 =2,
If and only if x = y = √ 2, the equal sign holds
So XY + 4 (x + y) - 2 ≤ 2 + 8 √ 2 - 2
=8√2.
When x = y = √ 2, XY + 4 (x + y) - 2 has a minimum value of 8 √ 2
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Mean inequality
That is, the square mean ≥ arithmetic mean ≥ geometric mean