Find the minimum value of ?^?+?^?+?^?
if ?^?+??^?+??^?=?? ; for real a, b & c
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10
12
12
8
For exam approach:
a=b=c=√2 satisfy 2nd equation.
So, 3*(√2)⁴ =12
12
a=b=c=√2
2+2*4+9*2=28
A^4+b^4+c^4=12
8
12
For minimum all the variable must have same value so a=b=c = ✓2
So answer is 12
12
12
A=2B=3C A=√2=B=C
12
12
12
12
8
28*28/98=8
12
8
8
Applying cauchy inequality (28)^2 <= 98 * (a^4+b^4+c^4)..
8 would be the minimum value
8
12 a=b=c
Ans is 8
From Cauchy–Schwarz inequality, (a^2 + 4b^2 + 9c^2)^2 <= (a^4+b^4+c^4)(1^2+4^2+9^2)
or, (a^4+b^4+c^4) * 98 >= 28^2 , or, (a^4+b^4+c^4) > = 8
Sir aapke pronunciation sahi nhi rehta pls improve
Sir, thank you for correcting me. a,b,c has to be real numbers for a qudratic equation with discriminant <0 to have roots as complex conjugates.
28*28/98 = 8