Difference between revisions of "1964 IMO Problems/Problem 2"

Latest revision as of 16:32, 4 July 2024

Problem

Suppose $a, b, c$ are the sides of a triangle. Prove that

$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}.$

Solution

Let $b+c-a = x$ , $c+a-b = y$ , and $a+b-c = z$ . Then, $a = \frac{y+z}{2}$ , $b = \frac{x+z}{2}$ , and $c = \frac{x+y}{2}$ . By AM-GM, $\frac{x+y}{2} \geq \sqrt{xy},$ $\frac{y+z}{2} \geq \sqrt{yz},$ $\textrm{and }\frac{x+z}{2} \geq \sqrt{xz}.$

Multiplying these equations, we have $\frac{x+y}{2} \cdot \frac{y+z}{2} \cdot \frac{x+z}{2} \geq xyz$ $\therefore abc \geq (a+b-c)(b+c-a)(c+a-b).$ We can now simplify: $(a+b-c)(b+c-a)(c+a-b) \leq abc$ $(-a^2 + b^2 - c^2 + 2ac)(c+a-b) \leq abc$ $a(-a^2 + b^2 - c^2 + 2ac) + c(-a^2 + b^2 - c^2 + 2ac) - b(-a^2 + b^2 - c^2 + 2ac) \leq abc$ $-a^3 + ab^2 - ac^2 + 2a^2c - a^2c + b^2c - c^3 + 2ac^2 + a^2b - b^3 + bc^2 - 2abc \leq abc$ $a^2b + a^2c - a^3 + b^2c + ab^2 - b^3 + ac^2 + bc^2 - c^3 - 2abc \leq abc$ $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}\textrm{. }\square$ ~mathboy100

Solution 2

We can use the substitution $a=x+y$ , $b=x+z$ , and $c=y+z$ to get

$2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\leq 3(x+y)(x+z)(y+z)$

$2zx^2+2zy^2+2yx^2+2yz^2+2xy^2+2xz^2+12xyz\leq 3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz$

$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq 6xyz$

$\frac{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}{6}\geq xyz=\sqrt[6]{x^2yx^2zy^2xy^2zz^2xz^2y}$

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 $a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \ge 0,$ which is true by Schur's inequality.

@@ Line 4: / Line 4: @@
 <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
+<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>\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}</cmath>
+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}}

1964 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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