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Let 
a,b,c > 0. Show that

(a+b+c)/(abc) <= (1)/(a^(2))+(1)/(b^(2))+(1)/(c^(2)).

Let a,b,c>0 a, b, c>0 . Show that\newlinea+b+cabc1a2+1b2+1c2. \frac{a+b+c}{a b c} \leq \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} .

Full solution

Q. Let a,b,c>0 a, b, c>0 . Show that\newlinea+b+cabc1a2+1b2+1c2. \frac{a+b+c}{a b c} \leq \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} .
  1. 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.\newlineSo, (1/a+1/b+1/c)/3(1/a×1/b×1/c)1/3(1/a + 1/b + 1/c)/3 \geq (1/a \times 1/b \times 1/c)^{1/3}.
  2. Simplify Right Side: Now, let's simplify the right side of the inequality.\newline(1a×1b×1c)13=(1abc)13(\frac{1}{a} \times \frac{1}{b} \times \frac{1}{c})^{\frac{1}{3}} = (\frac{1}{abc})^{\frac{1}{3}}.
  3. Raise to Power of 33: Raise both sides of the inequality to the power of 33 to get rid of the cube root. (1/a+1/b+1/c3)31abc\left(\frac{1/a + 1/b + 1/c}{3}\right)^3 \geq \frac{1}{abc}.
  4. Multiply by 2727: Multiplying both sides of the inequality by 2727 (which is 333^3) gives us:\newline(1a+1b+1c)327abc.(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^3 \geq \frac{27}{abc}.
  5. Expand Using Binomial Theorem: Now, we'll expand the left side of the inequality using the binomial theorem. \newline(1/a2+1/b2+1/c2+2(1/ab+1/ac+1/bc))327/(abc)(1/a^2 + 1/b^2 + 1/c^2 + 2\cdot(1/ab + 1/ac + 1/bc))^3 \geq 27/(abc).\newlineOops, this is incorrect. We should have expanded (1/a+1/b+1/c)3(1/a + 1/b + 1/c)^3, not just the squared terms.

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