Difference between revisions of "1974 USAMO Problems/Problem 2"

Latest revision as of 09:20, 24 August 2024

Problem

Prove that if $a$ , $b$ , and $c$ are positive real numbers, then

$a^ab^bc^c\ge (abc)^{(a+b+c)/3}$

Solution 1

Consider the function $f(x)=x\ln{x}$ . $f''(x)=\frac{1}{x}>0$ for $x>0$ ; therefore, it is a convex function and we can apply Jensen's Inequality:

$\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\frac{a+b+c}{3}\right)$

Apply AM-GM to get

$\frac{a+b+c}{3}\ge \sqrt[3]{abc}$

which implies

$\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\sqrt[3]{abc}\right)$

Rearranging,

$a\ln{a}+b\ln{b}+c\ln{c}\ge\left(\frac{a+b+c}{3}\right)\ln\left(abc\right)$

Because $f(x) = e^x$ is an increasing function, we can conclude that:

$e^{a\ln{a}+b\ln{b}+c\ln{c}}\ge{e}^{\ln\left(abc\right)(a+b+c)/3}$

which simplifies to the desired inequality.

Solution 2

Note that $(a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3$ .

So if we can prove that $a^ab^bc^c\ge a^bb^cc^a$ and $a^ab^bc^c\ge a^cb^ac^b$ , then we are done.

WLOG let $a\ge b\ge c$ .

Note that $(a^ab^bc^c)\cdot \left(\dfrac{c}{a}\right)^{a-b}\cdot \left(\dfrac{c}{b}\right)^{b-c}=a^bb^cc^a$ . Since $\dfrac{c}{a} \le 1$ , $\dfrac{c}{b} \le 1$ , $a-b \ge 0$ , and $b-c \ge 0$ , it follows that $a^ab^bc^c \ge a^bb^cc^a$ .

Note that $(a^ab^bc^c)\cdot \left(\dfrac{b}{a}\right)^{a-b}\cdot \left(\dfrac{c}{a}\right)^{b-c}=a^cb^ac^b$ . Since $\dfrac{b}{a} \le 1$ , $\dfrac{c}{a} \le 1$ , $a-b \ge 0$ , and $b-c \ge 0$ , it follows that $a^ab^bc^c \ge a^cb^ac^b$ .

Thus, $(a^ab^bc^c)^3\ge (a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3$ , and cube-rooting both sides gives $a^ab^bc^c\ge (abc)^{(a+b+c)/3}$ as desired.

Solution 3

WLOG let $a\ge b\ge c$ . Let $b = ax$ and $c = ay$ , where $x \ge 1$ and $y \ge 1$ .

We want to prove that $(a)^{a}(ax)^{ax}(ay)^{ay} \ge (a \cdot ax \cdot ay)^{\frac{a + ax + ay}{3}}$ .

Simplifying and combining terms on each side, we get $a^{a + ax + ay}x^{ax}y^{ay} \ge a^{a + ax + ay}(xy)^{\frac{a + ax + ay}{3}}$ .

Since $a > 0$ , we can divide out $a^{a + ax + ay}$ to get $x^{ax}y^{ay} \ge (xy)^{\frac{a + ax + ay}{3}}$ .

Take the $a$ th root of each side and then cube both sides to get $x^{3x}y^{3y} \ge (xy)^{1 + x + y}$ .

This simplifies to $x^{2x-1}y^{2y-1} \ge x^{y}y^{x}$ .

Since $2x - 1 \ge x$ and $2y - 1 \ge y$ , we only need to prove $x^{x}y^{y} \ge x^{y}y^{x}$ for our given $x, y$ .

WLOG, let $y \ge x$ and $y =kx$ for $k \ge 1$ . Then our expression becomes

$x^{x}(xk)^{xk} \ge x^{xk}(xk)^{x}$

$x^{x+xk}k^{xk} \ge x^{x+xk}k^{x}$

$k^{xk} \ge k^{x}$

$k^k \ge k$

This is clearly true for $k \ge 1$ .

Solution 4

WLOG let $a\ge b\ge c$ . Then sequence $(a,b,c)$ majorizes $(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3})$ . Thus by Muirhead's Inequality, we have $\sum_{sym} a^ab^bc^c \ge \sum_{sym} a^{\frac{a+b+c}{3}}b^{\frac{a+b+c}{3}}c^{\frac{a+b+c}{3}}$ , so $a^ab^bc^c \ge (abc)^{\frac{a+b+c}{3}}$ .

Solution 5

Let $x=\frac{a}{\sqrt[3]{abc}},$ $y=\frac{b}{\sqrt[3]{abc}}$ and $z=\frac{c}{\sqrt[3]{abc}}.$ Then $xyz=1$ and a straightforward calculation reduces the problem to $x^xy^yz^z \ge 1.$ WLOG, assume $x\ge y\ge z.$ Then $x\ge 1,$ $z\le 1$ and $xy=\frac{1}{z} \ge 1.$ Therefore, $x^xy^yz^z=x^{x-y}(xy)^{y-z}(xyz)^z \ge 1.$

J.Z.

Solution 6

Cubing both sides of the given inequality gives $a^{3a}b^{3b}c^{3c}\ge(abc)^{a+b+c}$ If we take $a^{3a}b^{3b}c^{3c}$ as the product of $3a$ $a$ 's, $3b$ $b$ 's, and $3c$ $c's$ , we get that

$\sqrt[3(a+b+c)]{a^{3a}b^{3b}c^{3c}}\ge\frac{3(a+b+c)}{\frac{3a}{a}+\frac{3b}{b}+\frac{3c}{c}}=\frac{a+b+c}{3}\ge\sqrt[3]{abc}$

by GM-HM, as desired.

Solution 7

Replacing $a$ with $ak$ , $b$ with $bk$ , and $c$ with $ck$ , for some positive real $k$ we get: $a^{ak}k^{ak}b^{bk}k^{bk}c^{ck}k^{ck} = a^{ak}b^{bk}c^{ck}k^{ak+bk+ck} \ge {k^3abc}^{{ak+bk+ck}/3} = k^{ak+bk+ck}{abc}^{{ak+bk+ck}/3}$ This means that this inequality is homogeneous since both sides have the same power of $k$ as a factor. Since the inequality is homogeneous, we can scale $abc$ so that their product is $1$ , i.e. $abc = 1$ . This makes the inequality turn into something much more nicer to deal with. Now we have to prove: $a^ab^bc^c \ge 1$ given that $abc = 1$ . Note that $a^a \ge a$ even if $a \le 1$ . Therefore $a^a \ge a$ , $b^b \ge b$ , and $c^c \ge c$ . Multiplying these together we get: $a^ab^bc^c \ge abc = 1$ . This proves the desired result. Equality holds when $a = b = c = 1$ .

Solution 8 (Rearrangement)

Let $\log_2a=x$ , $\log_2b=y$ and $\log_2c=z$ , and WLOG $a\ge b\ge c >0$ . Then we have both $a\ge b\ge c$ and $x\ge y \ge z$ . By the rearrangement inequality, $ax+by+cz\ge ay + bz +cx$ and $ax+by +cz\ge az+bx+cy$ . Summing, $2(ax+by+cz)\ge a(y+z)+b(x+z)+c(x+y)$ Adding $ax+by+cz$ ,we get $ax+by+cz\ge \frac{(a+b+c)(x+y+z)}{3}$ . Now we substitute back for $x,y,z$ to get: $\log_2a^ab^bc^c\ge \log_2(abc)^{\frac{a+b+c}{3}}$ . Raising $2$ to the power of each side, we get $a^ab^bc^c\ge (abc)^{\frac{a+b+c}{3}}$

Solution 9

Assume without loss of generality that $a\le b\le c$ . Then, we have $b/a\ge 1, c/a\ge 1, c/b\ge 1, b-a\ge 0, c-a\ge 0, c-b\ge 0.$ It follows that $(b/a)^{(b-a)/3}\cdot (c/a)^{(c-a)/3}\cdot (c/b)^{(c-b)/3}\ge 1.$ We can "distribute" the exponents, which gives $\dfrac{b^{(b-a)/3}}{a^{(b-a)/3}}\cdot \dfrac{c^{(c-a)/3}}{a^{(c-a)/3}}\cdot \dfrac{c^{(c-b)/3}}{b^{(c-b)/3}}\ge 1.$ Notice that in the $b^{(c-b)/3}$ in the denominator of the third fraction can be brought out of the denominator by negating its exponent. This gives $\dfrac{b^{(b-a)/3}}{a^{(b-a)/3}}\cdot \dfrac{c^{(c-a)/3}}{a^{(c-a)/3}}\cdot c^{(c-b)/3} \cdot b^{(b-c)/3}\ge 1.$ We can move around terms on the left to get $\left(\dfrac{b^{(b-a)/3}\cdot b^{(b-c)/3}}{a^{(b-a)/3}}\right)\left(\dfrac{c^{(c-a)/3}\cdot c^{(c-b)/3}}{a^{(c-a)/3}}\right)\ge 1.$ Now we combine the exponents in the numerators. This gives $\left(\dfrac{b^{(2b-c-a)/3}}{a^{(b-a)/3}}\right)\left(\dfrac{c^{(2c-b-a)/3}}{a^{(c-a)/3}}\right)\ge 1.$ We multiply both sides by $\left(a^{(b-a)/3}\right)\left(a^{(c-a)/3}\right),$ which gives $\left(b^{(2b-c-a)/3}\right)\left(c^{(2c-b-a)/3}\right)\ge \left(a^{(b-a)/3}\right)\left(a^{(c-a)/3}\right).$ We then multiply both sides by $\left(b^{(c-a)/3}\right)\left(c^{(b-a)/3}\right):$ $\left(b^{(2b-2a)/3}\right)\left(c^{(2c-2a)/3}\right)\ge \left(a^{(b-a)/3}\right)\left(c^{(b-a)/3}\right)\left(a^{(c-a)/3}\right)\left(b^{(c-a)/3}\right).$ Multiplying both sides by $\left(b^{(b-a)/3}\right)\left(c^{(c-a)/3}\right),$ then simplifying both sides with exponent laws, gives $b^{b-a}\cdot c^{c-a}\ge (abc)^{(b+c-2a)/3}.$ Finally, we multiply both sides by $(abc)^a.$ Combining exponents one last time gives the desired $a^a b^b c^c\ge (abc)^{(a+b+c)/3}.$ --MenuThreeOne

Video Solution: Only basic rules of exponents are needed

https://youtu.be/lXmUugsPb1U?si=JAp8PJ5DxnULtUTB

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

@@ Line 3: / Line 3: @@
 <center><math>a^ab^bc^c\ge (abc)^{(a+b+c)/3}</math></center>
-==Solution==
+==Solution 1==
 Consider the function <math>f(x)=x\ln{x}</math>. <math>f''(x)=\frac{1}{x}>0</math> for <math>x>0</math>; therefore, it is a convex function and we can apply [[Jensen's Inequality]]:
 <center><math>\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\frac{a+b+c}{3}\right)</math></center>
@@ Line 15: / Line 15: @@
 <center><math>e^{a\ln{a}+b\ln{b}+c\ln{c}}\ge{e}^{\ln\left(abc\right)(a+b+c)/3}</math></center>
 which simplifies to the desired inequality.
+==Solution 2==
+Note that <math>(a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3</math>.
+So if we can prove that <math>a^ab^bc^c\ge a^bb^cc^a</math> and <math>a^ab^bc^c\ge a^cb^ac^b</math>, then we are done.
+WLOG let <math>a\ge b\ge c</math>.
+Note that <math>(a^ab^bc^c)\cdot \left(\dfrac{c}{a}\right)^{a-b}\cdot \left(\dfrac{c}{b}\right)^{b-c}=a^bb^cc^a</math>. Since <math>\dfrac{c}{a} \le  1</math>, <math>\dfrac{c}{b} \le  1</math>, <math>a-b \ge 0</math>, and <math>b-c \ge 0</math>, it follows that <math>a^ab^bc^c \ge a^bb^cc^a</math>.
+Note that <math>(a^ab^bc^c)\cdot \left(\dfrac{b}{a}\right)^{a-b}\cdot \left(\dfrac{c}{a}\right)^{b-c}=a^cb^ac^b</math>. Since <math>\dfrac{b}{a} \le  1</math>, <math>\dfrac{c}{a} \le  1</math>, <math>a-b \ge 0</math>, and <math>b-c \ge 0</math>, it follows that <math>a^ab^bc^c \ge a^cb^ac^b</math>.
+Thus, <math>(a^ab^bc^c)^3\ge (a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3</math>, and cube-rooting both sides gives <math>a^ab^bc^c\ge (abc)^{(a+b+c)/3}</math> as desired.
+==Solution 3==
+WLOG let <math>a\ge b\ge c</math>. Let <math>b = ax</math> and <math>c = ay</math>, where <math>x \ge 1</math> and <math>y \ge 1</math>.
+We want to prove that <math>(a)^{a}(ax)^{ax}(ay)^{ay} \ge (a \cdot ax \cdot ay)^{\frac{a + ax + ay}{3}}</math>.
+Simplifying and combining terms on each side, we get <math>a^{a + ax + ay}x^{ax}y^{ay} \ge a^{a + ax + ay}(xy)^{\frac{a + ax + ay}{3}}</math>.
+Since <math>a > 0</math>, we can divide out <math>a^{a + ax + ay}</math> to get <math>x^{ax}y^{ay} \ge (xy)^{\frac{a + ax + ay}{3}}</math>.
+Take the <math>a</math>th root of each side and then cube both sides to get <math>x^{3x}y^{3y} \ge (xy)^{1 + x + y}</math>.
+This simplifies to <math>x^{2x-1}y^{2y-1} \ge x^{y}y^{x}</math>.
+Since <math>2x - 1 \ge x</math> and <math>2y - 1 \ge y</math>, we only need to prove <math>x^{x}y^{y} \ge x^{y}y^{x}</math> for our given <math>x, y</math>.
+WLOG, let <math>y \ge x</math> and <math> y =kx</math> for <math>k \ge 1</math>. Then our expression becomes
+<math>x^{x}(xk)^{xk} \ge x^{xk}(xk)^{x}</math>
+<math>x^{x+xk}k^{xk} \ge x^{x+xk}k^{x}</math>
+<math>k^{xk} \ge k^{x}</math>
+<math>k^k \ge k</math>
+This is clearly true for <math>k \ge 1</math>.
+==Solution 4==
+WLOG let <math>a\ge b\ge c</math>. Then sequence <math>(a,b,c)</math> majorizes <math>(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3})</math>. Thus by Muirhead's Inequality, we have <math>\sum_{sym} a^ab^bc^c \ge \sum_{sym} a^{\frac{a+b+c}{3}}b^{\frac{a+b+c}{3}}c^{\frac{a+b+c}{3}}</math>, so <math>a^ab^bc^c \ge (abc)^{\frac{a+b+c}{3}}</math>.
+==Solution 5==
+Let <math>x=\frac{a}{\sqrt[3]{abc}},</math> <math>y=\frac{b}{\sqrt[3]{abc}}</math> and <math>z=\frac{c}{\sqrt[3]{abc}}.</math> Then <math>xyz=1</math> and a straightforward calculation reduces the problem to
+<cmath>x^xy^yz^z \ge 1.</cmath>
+WLOG, assume <math>x\ge y\ge z.</math> Then <math>x\ge 1,</math> <math>z\le 1</math> and <math>xy=\frac{1}{z} \ge 1.</math> Therefore,
+<cmath> x^xy^yz^z=x^{x-y}(xy)^{y-z}(xyz)^z \ge 1.</cmath>
+J.Z.
+==Solution 6==
+Cubing both sides of the given inequality gives <cmath>a^{3a}b^{3b}c^{3c}\ge(abc)^{a+b+c}</cmath>
+If we take <math>a^{3a}b^{3b}c^{3c}</math> as the product of <math>3a</math> <math>a</math>'s, <math>3b</math> <math>b</math>'s, and <math>3c</math> <math>c's</math>, we get that
+<center>
+<math>\sqrt[3(a+b+c)]{a^{3a}b^{3b}c^{3c}}\ge\frac{3(a+b+c)}{\frac{3a}{a}+\frac{3b}{b}+\frac{3c}{c}}=\frac{a+b+c}{3}\ge\sqrt[3]{abc}</math>
+</center>
+by GM-HM, as desired.
+==Solution 7==
+Replacing <math>a</math> with <math>ak</math>, <math>b</math> with <math>bk</math>, and <math>c</math> with <math>ck</math>, for some positive real <math>k</math> we get:
+<math>a^{ak}k^{ak}b^{bk}k^{bk}c^{ck}k^{ck} = a^{ak}b^{bk}c^{ck}k^{ak+bk+ck} \ge {k^3abc}^{{ak+bk+ck}/3} = k^{ak+bk+ck}{abc}^{{ak+bk+ck}/3}</math>
+This means that this inequality is homogeneous since both sides have the same power of <math>k</math> as a factor. Since the inequality is homogeneous, we can scale <math>abc</math> so that their product is <math>1</math>, i.e. <math>abc = 1</math>. This makes the inequality turn into something much more nicer to deal with. Now we have to prove:
+<math>a^ab^bc^c \ge 1</math> given that <math>abc = 1</math>.
+Note that <math>a^a \ge a</math> even if <math>a \le 1</math>. Therefore <math>a^a \ge a</math>, <math>b^b \ge b</math>, and <math>c^c \ge c</math>. Multiplying these together we get:
+<math>a^ab^bc^c \ge abc = 1</math>. This proves the desired result. Equality holds when <math>a = b = c = 1</math>.
+==Solution 8 (Rearrangement)==
+Let <math>\log_2a=x</math>, <math>\log_2b=y</math> and <math>\log_2c=z</math>, and WLOG <math>a\ge b\ge c >0</math>.
+Then we have both <math>a\ge b\ge c</math> and <math>x\ge y \ge z</math>.
+By the rearrangement inequality,
+<math>ax+by+cz\ge ay + bz +cx</math> and
+<math>ax+by +cz\ge az+bx+cy</math>.
+Summing,
+<math>2(ax+by+cz)\ge a(y+z)+b(x+z)+c(x+y)</math>
+Adding <math>ax+by+cz</math>,we get
+<math>ax+by+cz\ge \frac{(a+b+c)(x+y+z)}{3}</math>.
+Now we substitute back for <math>x,y,z</math> to get:
+<math>\log_2a^ab^bc^c\ge \log_2(abc)^{\frac{a+b+c}{3}}</math>.
+Raising <math>2</math> to the power of each side, we get
+<math>a^ab^bc^c\ge (abc)^{\frac{a+b+c}{3}}</math>
+==Solution 9==
+Assume without loss of generality that <math>a\le b\le c</math>. Then, we have
+<cmath>b/a\ge 1, c/a\ge 1, c/b\ge 1, b-a\ge 0, c-a\ge 0, c-b\ge 0.</cmath>
+It follows that
+<cmath>(b/a)^{(b-a)/3}\cdot (c/a)^{(c-a)/3}\cdot (c/b)^{(c-b)/3}\ge 1.</cmath>
+We can "distribute" the exponents, which gives
+<cmath>\dfrac{b^{(b-a)/3}}{a^{(b-a)/3}}\cdot \dfrac{c^{(c-a)/3}}{a^{(c-a)/3}}\cdot \dfrac{c^{(c-b)/3}}{b^{(c-b)/3}}\ge 1.</cmath>
+Notice that in the <math>b^{(c-b)/3}</math> in the denominator of the third fraction can be brought out of the denominator by negating its exponent. This gives
+<cmath>\dfrac{b^{(b-a)/3}}{a^{(b-a)/3}}\cdot \dfrac{c^{(c-a)/3}}{a^{(c-a)/3}}\cdot c^{(c-b)/3} \cdot b^{(b-c)/3}\ge 1.</cmath>
+We can move around terms on the left to get
+<cmath>\left(\dfrac{b^{(b-a)/3}\cdot b^{(b-c)/3}}{a^{(b-a)/3}}\right)\left(\dfrac{c^{(c-a)/3}\cdot c^{(c-b)/3}}{a^{(c-a)/3}}\right)\ge 1.</cmath>
+Now we combine the exponents in the numerators. This gives
+<cmath>\left(\dfrac{b^{(2b-c-a)/3}}{a^{(b-a)/3}}\right)\left(\dfrac{c^{(2c-b-a)/3}}{a^{(c-a)/3}}\right)\ge 1.</cmath>
+We multiply both sides by <math>\left(a^{(b-a)/3}\right)\left(a^{(c-a)/3}\right),</math> which gives
+<cmath>\left(b^{(2b-c-a)/3}\right)\left(c^{(2c-b-a)/3}\right)\ge \left(a^{(b-a)/3}\right)\left(a^{(c-a)/3}\right).</cmath>
+We then multiply both sides by <math>\left(b^{(c-a)/3}\right)\left(c^{(b-a)/3}\right):</math>
+<cmath>\left(b^{(2b-2a)/3}\right)\left(c^{(2c-2a)/3}\right)\ge \left(a^{(b-a)/3}\right)\left(c^{(b-a)/3}\right)\left(a^{(c-a)/3}\right)\left(b^{(c-a)/3}\right).</cmath>
+Multiplying both sides by <math>\left(b^{(b-a)/3}\right)\left(c^{(c-a)/3}\right),</math> then simplifying both sides with exponent laws, gives
+<cmath>b^{b-a}\cdot c^{c-a}\ge (abc)^{(b+c-2a)/3}.</cmath>
+Finally, we multiply both sides by <math>(abc)^a.</math> Combining exponents one last time gives the desired
+<cmath>a^a b^b c^c\ge (abc)^{(a+b+c)/3}.</cmath>
+--MenuThreeOne
+==Video Solution: Only basic rules of exponents are needed==
+https://youtu.be/lXmUugsPb1U?si=JAp8PJ5DxnULtUTB
 {{alternate solutions}}
-== Mathlinks Discussions ==
+== See Also ==
+{{USAMO box|year=1974|num-b=1|num-a=3}}
 *[http://www.mathlinks.ro/viewtopic.php?t=102633 Simple Olympiad Inequality]
 *[http://www.mathlinks.ro/viewtopic.php?t=98846 Hard inequality]
@@ Line 25: / Line 142: @@
 *[http://www.mathlinks.ro/viewtopic.php?t=213258 ineq]
 *[http://www.mathlinks.ro/Forum/viewtopic.php?t=46247 exponents (generalization)]
+{{MAA Notice}}
-{{USAMO box|year=1974|num-b=1|num-a=3}}
 [[Category:Olympiad Algebra Problems]]
 [[Category:Olympiad Inequality Problems]]

1974 USAMO (Problems • Resources)
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