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Difference between revisions of "1995 IMO Problems/Problem 2"

(→‎Solution 3)
(→‎Solution 3b)
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Dividing by 2(ab + bc + ac) and noting that <math>ab + bc + ac \ge 3(a^2b^2c^2)^{\frac{1}{3}} = 3</math> by AM-GM gives
 
Dividing by 2(ab + bc + ac) and noting that <math>ab + bc + ac \ge 3(a^2b^2c^2)^{\frac{1}{3}} = 3</math> by AM-GM gives
−
<cmath>(frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)} \ge \frac{ab + bc + ac}{2} \ge \frac{3}{2},</cmath>
+
<cmath>\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)} \ge \frac{ab + bc + ac}{2} \ge \frac{3}{2},</cmath>
 
as desired.
 
as desired.
  

Revision as of 22:18, 21 April 2014

Problem

(Nazar Agakhanov, Russia) Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that \[\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}.\]

Solution

Solution 1

We make the substitution $x= 1/a$, $y=1/b$, $z=1/c$. Then \begin{align*} \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} &= \frac{x^3}{xyz(1/y+1/z)} + \frac{y^3}{xyz(1/z+1/x)} + \frac{z^3}{xyz(1/x+1/z)} \\ &= \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} . \end{align*} Since $(x^2,y^2,z^2)$ and $\bigl( 1/(y+z), 1/(z+x), 1/(x+y) \bigr)$ are similarly sorted sequences, it follows from the Rearrangement Inequality that \[\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{1}{2} \left( \frac{y^2+z^2}{y+z} + \frac{z^2+x^2}{z+x} + \frac{x^2+y^2}{x+y} \right) .\] By the Power Mean Inequality, \[\frac{y^2+z^2}{y+z} \ge \frac{(y+z)^2}{2(x+y)} = \frac{x+y}{2} .\] Symmetric application of this argument yields \[\frac{1}{2}\left( \frac{y^2+z^2}{y+z} + \frac{z^2+x^2}{z+x} + \frac{x^2+y^2}{x+y} \right) \ge \frac{1}{2}(x+y+z) .\] Finally, AM-GM gives us \[\frac{1}{2}(x+y+z) \ge \frac{3}{2}xyz = \frac{3}{2},\] as desired. $\blacksquare$

Solution 2

We make the same substitution as in the first solution. We note that in general, \[\frac{p}{q+r} = \frac{(p+q+r)}{(p+q+r)-p} - 1 .\] It follows that $(x,y,z)$ and $\bigl(x/(y+z), y/(z+x), z/(x+y)\bigr)$ are similarly sorted sequences. Then by Chebyshev's Inequality, \[\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{1}{3}(x+y+z) \left(\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} \right) .\] By AM-GM, $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}=1$, and by Nesbitt's Inequality, \[\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} \ge \frac{3}{2}.\] The desired conclusion follows. $\blacksquare$

Solution 3

Without clever substitutions: By Cauchy-Schwarz, \[\left(\sum_{cyc}\dfrac{1}{a^3 (b+c)}\right)\left(\sum_{cyc}a(b+c)\right)\geq \left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right) ^2=(ab+ac+bc)^2\] Dividing by $2(ab+bc+ac)$ gives \[\dfrac{1}{a^3 (b+c)}+\dfrac{1}{b^3 (a+c)}+\dfrac{1}{c^3 (a+b)}\geq \dfrac{1}{2}(ab+bc+ac)\geq \dfrac{3}{2}\] by AM-GM.

Solution 3b

Without clever notation: By Cauchy-Schwarz, \[(a(b+c) + b(c+a) + c(a+b)) \cdot (\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)})\] \[\ge (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2\] \[= (ab + bc + ac)^2\]

Dividing by 2(ab + bc + ac) and noting that $ab + bc + ac \ge 3(a^2b^2c^2)^{\frac{1}{3}} = 3$ by AM-GM gives \[\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)} \ge \frac{ab + bc + ac}{2} \ge \frac{3}{2},\] as desired.

Solution 4

Proceed as in Solution 1, to arrive at the equivalent inequality \[\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2} .\] But we know that \[x + y + z \ge 3xyz \ge 3\] by AM-GM. Furthermore, \[(x + y + y + z + x + z) (\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y}) \ge (x + y + z)^2\] by Cauchy-Schwarz, and so dividing by $2(x + y + z)$ gives \begin{align*}\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} &\ge \frac{(x + y + z)}{2} \\ &\ge \frac{3}{2} \end{align*}, as desired.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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