Difference between revisions of "1995 IMO Problems/Problem 2"
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Note that <math>abc = 1 \implies a = \frac{1}{bc}</math>. The cyclic sum becomes <math>\sum_{cyc}\frac{(bc)^3}{b + c}</math>. Note that by AM-GM, the cyclic sum is greater than or equal to <math>3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math>. We now see that we have the three so we must be on the right path. We now only need to show that <math>\frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^\frac13</math>. Notice that by AM-GM, <math>a + b \geq 2\sqrt{ab}</math>, <math>b + c \geq 2\sqrt{bc}</math>, and <math>a + c \geq 2\sqrt{ac}</math>. Thus, we see that <math>(a+b)(b+c)(a+c) \geq 8</math>, concluding that <math>\sum_{cyc} \frac{(bc)^3}{b+c} \geq \frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math> | Note that <math>abc = 1 \implies a = \frac{1}{bc}</math>. The cyclic sum becomes <math>\sum_{cyc}\frac{(bc)^3}{b + c}</math>. Note that by AM-GM, the cyclic sum is greater than or equal to <math>3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math>. We now see that we have the three so we must be on the right path. We now only need to show that <math>\frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^\frac13</math>. Notice that by AM-GM, <math>a + b \geq 2\sqrt{ab}</math>, <math>b + c \geq 2\sqrt{bc}</math>, and <math>a + c \geq 2\sqrt{ac}</math>. Thus, we see that <math>(a+b)(b+c)(a+c) \geq 8</math>, concluding that <math>\sum_{cyc} \frac{(bc)^3}{b+c} \geq \frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math> | ||
+ | === Solution 6 === | ||
+ | We want to try and apply Cauchy (Titu), by transforming the numerator into a quadratic expression and the denominator into a linear expression. This is easily achieved by using the provided condition: \ | ||
− | === Solution | + | Since <math>abc=1, \frac{1}{a^3(b+c)}=\frac{a^2b^2c^2}{a^3(b+c)}=\frac{b^2c^2}{a(b+c)}</math>. Likewise, <math>\frac{1}{b^3(c+a)}=\frac{a^2c^2}{b(a+c)},</math> and <math>\frac{1}{c^3(a+b)}=\frac{a^2b^2}{c(a+b)}</math>. Hence, by Cauchy (Titu): |
+ | |||
+ | \begin{align*} | ||
+ | \sum_{\text{cyc}}\frac{1}{a^3(b+c)} & \ge \frac{(bc+ac+ab)^2}{a(b+c)+b(a+c)+c(a+b)} \& = \frac{(bc+ac+ab)^2}{2bc+2ac+2ab} \& = \frac{bc+ac+ab}{2} \& \ge \frac{3\sqrt[^3]{a^2b^2c^2}}{2}, \text{by AM-GM} \& = \frac{3}{2}, \text{by the given condition <math>abc=1</math>}. | ||
+ | |||
+ | === Solution 7 from Brilliant Wiki (Muirheads) ==== | ||
https://brilliant.org/wiki/muirhead-inequality/ | https://brilliant.org/wiki/muirhead-inequality/ | ||
Revision as of 09:43, 20 June 2020
Contents
[hide]Problem
(Nazar Agakhanov, Russia) Let be positive real numbers such that . Prove that
Solution
Solution 1
We make the substitution , , . Then Since and are similarly sorted sequences, it follows from the Rearrangement Inequality that By the Power Mean Inequality, Symmetric application of this argument yields Finally, AM-GM gives us as desired.
Solution 2
We make the same substitution as in the first solution. We note that in general, It follows that and are similarly sorted sequences. Then by Chebyshev's Inequality, By AM-GM, , and by Nesbitt's Inequality, The desired conclusion follows.
Solution 3
Without clever substitutions: By Cauchy-Schwarz, Dividing by gives by AM-GM.
Solution 3b
Without clever notation: By Cauchy-Schwarz,
Dividing by and noting that by AM-GM gives as desired.
Solution 4
Proceed as in Solution 1, to arrive at the equivalent inequality But we know that by AM-GM. Furthermore, by Cauchy-Schwarz, and so dividing by gives as desired.
Solution 5
Without clever substitutions, and only AM-GM!
Note that . The cyclic sum becomes . Note that by AM-GM, the cyclic sum is greater than or equal to . We now see that we have the three so we must be on the right path. We now only need to show that . Notice that by AM-GM, , , and . Thus, we see that , concluding that
Solution 6
We want to try and apply Cauchy (Titu), by transforming the numerator into a quadratic expression and the denominator into a linear expression. This is easily achieved by using the provided condition: \
Since . Likewise, and . Hence, by Cauchy (Titu):
\begin{align*}
\sum_{\text{cyc}}\frac{1}{a^3(b+c)} & \ge \frac{(bc+ac+ab)^2}{a(b+c)+b(a+c)+c(a+b)} \\& = \frac{(bc+ac+ab)^2}{2bc+2ac+2ab} \\& = \frac{bc+ac+ab}{2} \\& \ge \frac{3\sqrt[^3]{a^2b^2c^2}}{2}, \text{by AM-GM} \\& = \frac{3}{2}, \text{by the given condition }.
Solution 7 from Brilliant Wiki (Muirheads) =
https://brilliant.org/wiki/muirhead-inequality/
Scroll all the way down Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.