Q. Let a,b,c>0. Show thatabca+b+c≤a21+b21+c21.
Apply AM-GM Inequality: We'll start by applying the AM-GM inequality, which states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean.So, (1/a+1/b+1/c)/3≥(1/a×1/b×1/c)1/3.
Simplify Right Side: Now, let's simplify the right side of the inequality.(a1×b1×c1)31=(abc1)31.
Raise to Power of 3: Raise both sides of the inequality to the power of 3 to get rid of the cube root. (31/a+1/b+1/c)3≥abc1.
Multiply by 27: Multiplying both sides of the inequality by 27 (which is 33) gives us:(a1+b1+c1)3≥abc27.
Expand Using Binomial Theorem: Now, we'll expand the left side of the inequality using the binomial theorem. (1/a2+1/b2+1/c2+2⋅(1/ab+1/ac+1/bc))3≥27/(abc).Oops, this is incorrect. We should have expanded (1/a+1/b+1/c)3, not just the squared terms.
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