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Finding the best value of basic inequality It is known that 4A ^ 2 + B ^ 2 = 6 Finding the minimum of a + B The known condition is changed to a ^ 2 + 2 (b ^ 2) = 6 The minimum value of a + B is - 3, which should be solved by basic inequality.
4A ^ 2 + B ^ 2 = 6 transform elliptic equation
a^2/(√6/2)^2+b^2/(√6)^2=1
X=√6/2*cosA/,y=√6sinA
X+Y=√6/2*cosA+√6sinA
=√6(1/2*cosA/+sinA)
Let tanb = 1 / 2, CoSb = 2 / √ 5
X+Y=√6/cosB*(sinB*cosA+sinA*cosB)
=√30/2*sin(A+B)
(X+Y)max=√30/2
(X+Y)min=0
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