i will mark brainliest, thank, and 5star Find the smallest positive value of the nonnegative number \lambdaλ such that the inequality
\frac{a+b}{2}\ge \lambda \sqrt{ab}+(1-\lambda) \sqrt{\frac{a^2+b^2}{2}}
2
a+b
​
≥λ
ab
​
+(1−λ)
2
a
2
+b
2

​

​

holds for all positive real numbers a, ba, b.

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step-by-step explanation:

add 5 subtract 1

step-by-step explanation:

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph