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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>
  
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=== 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: \\
  
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=== Solution 6 from Brilliant Wiki (Muirheads) ====
+
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*}
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    \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>}.
 +
 
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=== 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

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, \[\left(a(b+c) + b(c+a) + c(a+b)\right) \cdot \left(\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)}\right)\] \[\ge \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^2\] \[= (ab + bc + ac)^2\]

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.

Solution 5

Without clever substitutions, and only AM-GM!

Note that $abc = 1 \implies a = \frac{1}{bc}$. The cyclic sum becomes $\sum_{cyc}\frac{(bc)^3}{b + c}$. Note that by AM-GM, the cyclic sum is greater than or equal to $3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}$. We now see that we have the three so we must be on the right path. We now only need to show that $\frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^\frac13$. Notice that by AM-GM, $a + b \geq 2\sqrt{ab}$, $b + c \geq 2\sqrt{bc}$, and $a + c \geq 2\sqrt{ac}$. Thus, we see that $(a+b)(b+c)(a+c) \geq 8$, concluding that $\sum_{cyc} \frac{(bc)^3}{b+c} \geq \frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}$

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 $abc=1, \frac{1}{a^3(b+c)}=\frac{a^2b^2c^2}{a^3(b+c)}=\frac{b^2c^2}{a(b+c)}$. Likewise, $\frac{1}{b^3(c+a)}=\frac{a^2c^2}{b(a+c)},$ and $\frac{1}{c^3(a+b)}=\frac{a^2b^2}{c(a+b)}$. 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 $abc=1$}.

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.

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