Question

a*2+b*2+c*2+d*2=1 then maximum value of abcd​

Answer 1

Answer:

a^2 +b^2 + c^2 +d^2 =1

Now we know AM>= GM >=HM

Therefore, arithmetic mean of a^2, b^2, c^2, d^2 is 1/4

(a^2 +b^2 + c^2 +d^2) /4 =1/4

Geometric mean of a^2, b^2, c^2, d^2 = (a^2 * b^2 * c^2 * d^2) ^1/4

i.e 1/4>= (a^2 * b^2 * c^2 * d^2) ^1/4

Squaring both sides, 1/16 > = (a^2 * b^2 * c^2 * d^2) ^1/2

i.e 1/16>= abcd

Therefore the maximum value of 16abcd is (1/16) *16 = 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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