Difference between revisions of "1964 IMO Problems/Problem 2"
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<cmath>a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}.</cmath> | <cmath>a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}.</cmath> | ||
| − | == Solution == | + | ==Solution== |
| + | Let <math>b+c-a = x</math>, <math>c+a-b = y</math>, and <math>a+b-c = z</math>. Then, <math>a = \frac{y+z}{2}</math>, <math>b = \frac{x+z}{2}</math>, and <math>c = \frac{x+y}{2}</math>. By AM-GM, | ||
| + | <cmath>\frac{x+y}{2} \geq \sqrt{xy}, </cmath> | ||
| + | <cmath>\frac{y+z}{2} \geq \sqrt{yz}, </cmath> | ||
| + | <cmath>\textrm{and }\frac{x+z}{2} \geq \sqrt{xz}.</cmath> | ||
| + | |||
| + | Multiplying these equations, we have | ||
| + | <cmath>\frac{x+y}{2} \cdot \frac{y+z}{2} \cdot \frac{x+z}{2} \geq xyz</cmath> | ||
| + | <cmath>\therefore abc \geq (a+b-c)(b+c-a)(c+a-b).</cmath> | ||
| + | We can now simplify: | ||
| + | <cmath>(a+b-c)(b+c-a)(c+a-b) \leq abc</cmath> | ||
| + | <cmath>(-a^2 + b^2 - c^2 + 2ac)(c+a-b) \leq abc</cmath> | ||
| + | <cmath>a(-a^2 + b^2 - c^2 + 2ac) + c(-a^2 + b^2 - c^2 + 2ac) - b(-a^2 + b^2 - c^2 + 2ac) \leq abc</cmath> | ||
| + | <cmath>-a^3 + ab^2 - ac^2 + 2a^2c - a^2c + b^2c - c^3 + 2ac^2 + a^2b - b^3 + bc^2 - 2abc \leq abc</cmath> | ||
| + | <cmath>a^2b + a^2c - a^3 + b^2c + ab^2 - b^3 + ac^2 + bc^2 - c^3 - 2abc \leq abc</cmath> | ||
| + | <cmath>a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}\textrm{. }\square</cmath> | ||
| + | ~mathboy100 | ||
| + | |||
| + | == Solution 2 == | ||
We can use the substitution <math>a=x+y</math>, <math>b=x+z</math>, and <math>c=y+z</math> to get | We can use the substitution <math>a=x+y</math>, <math>b=x+z</math>, and <math>c=y+z</math> to get | ||
<cmath>2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\leq 3(x+y)(x+z)(y+z)</cmath> | <cmath>2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\leq 3(x+y)(x+z)(y+z)</cmath> | ||
| − | < | + | <math>2zx^2+2zy^2+2yx^2+2yz^2+2xy^2+2xz^2+12xyz\leq 3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz</math> |
<cmath>x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq 6xyz</cmath> | <cmath>x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq 6xyz</cmath> | ||
| Line 16: | Line 34: | ||
This is true by AM-GM. We can work backwards to get that the original inequality is true. | This is true by AM-GM. We can work backwards to get that the original inequality is true. | ||
| + | |||
| + | eevee9406: This solution uses incorrect reasoning regarding the inequalities. | ||
| + | |||
| + | ==Solution 3== | ||
| + | Rearrange to get | ||
| + | <cmath>a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \ge 0,</cmath> | ||
| + | which is true by Schur's inequality. | ||
| + | |||
| + | == See Also == {{IMO box|year=1964|num-b=1|num-a=3}} | ||
Latest revision as of 16:32, 4 July 2024
Problem
Suppose 
are the sides of a triangle. Prove that
![\[a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}.\]](http://latex.artofproblemsolving.com/2/c/8/2c8afb405fe890dd89759ea3a2ed2e527038adfc.png)
Solution
Let 
, 
, and 
. Then, 
, 
, and 
. By AM-GM,
![\[\frac{x+y}{2} \geq \sqrt{xy},\]](http://latex.artofproblemsolving.com/d/b/b/dbbbec16dcb8a296b18f0f27af022d91a5b8c356.png)
![\[\frac{y+z}{2} \geq \sqrt{yz},\]](http://latex.artofproblemsolving.com/c/f/e/cfed5cdca9b7fd9834c947d1a7021ee5df21d500.png)
![\[\textrm{and }\frac{x+z}{2} \geq \sqrt{xz}.\]](http://latex.artofproblemsolving.com/2/f/2/2f2846bf665638e8b929129491bf2d813cc40b11.png)
Multiplying these equations, we have
![\[\frac{x+y}{2} \cdot \frac{y+z}{2} \cdot \frac{x+z}{2} \geq xyz\]](http://latex.artofproblemsolving.com/7/a/4/7a4e4c7005b57bb08004083175b3dd3a3d6fc9ae.png)
![\[\therefore abc \geq (a+b-c)(b+c-a)(c+a-b).\]](http://latex.artofproblemsolving.com/f/0/a/f0ae2bd3e6a36696d11d095be34fd7f5d54c8164.png)
We can now simplify:
![\[(a+b-c)(b+c-a)(c+a-b) \leq abc\]](http://latex.artofproblemsolving.com/7/5/0/7505abb3b85d47c1c911b3a512f7dbea09edef79.png)
![\[(-a^2 + b^2 - c^2 + 2ac)(c+a-b) \leq abc\]](http://latex.artofproblemsolving.com/5/4/a/54ae25913870858066cf7ea7db8b52b0986240a7.png)
![\[a(-a^2 + b^2 - c^2 + 2ac) + c(-a^2 + b^2 - c^2 + 2ac) - b(-a^2 + b^2 - c^2 + 2ac) \leq abc\]](http://latex.artofproblemsolving.com/5/0/b/50bd407289b1dfc13a02a29cf680b02b7b4e97e4.png)
![\[-a^3 + ab^2 - ac^2 + 2a^2c - a^2c + b^2c - c^3 + 2ac^2 + a^2b - b^3 + bc^2 - 2abc \leq abc\]](http://latex.artofproblemsolving.com/0/1/f/01f17474eaebc642cbe91ebcde5e79464aa95046.png)
![\[a^2b + a^2c - a^3 + b^2c + ab^2 - b^3 + ac^2 + bc^2 - c^3 - 2abc \leq abc\]](http://latex.artofproblemsolving.com/b/7/a/b7acb76cb4da1cbd5c46ae648f782c8a8cc75285.png)
![\[a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}\textrm{. }\square\]](http://latex.artofproblemsolving.com/9/c/2/9c2073b73bee2f0ec5cab0a7604ce5492bab776b.png)
~mathboy100
Solution 2
We can use the substitution 
, 
, and 
to get
![\[2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\leq 3(x+y)(x+z)(y+z)\]](http://latex.artofproblemsolving.com/6/5/2/6522426f4f40387f35379270676d7bcd6ef21add.png)
![\[x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq 6xyz\]](http://latex.artofproblemsolving.com/4/4/c/44c76b06c472625ded2773717bcfa933af468183.png)
![\[\frac{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}{6}\geq xyz=\sqrt[6]{x^2yx^2zy^2xy^2zz^2xz^2y}\]](http://latex.artofproblemsolving.com/d/7/c/d7cdaef5d710f396683d10490561a1d321effebb.png)
This is true by AM-GM. We can work backwards to get that the original inequality is true.
eevee9406: This solution uses incorrect reasoning regarding the inequalities.
Solution 3
Rearrange to get
![\[a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \ge 0,\]](http://latex.artofproblemsolving.com/0/d/c/0dcd3e5fa64e8656215b2e2662a9f90f74cd08cd.png)
which is true by Schur's inequality.
See Also
| 1964 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
