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

Latest revision as of 17:34, 4 September 2025

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 (Ravi Substitution)

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.

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.

Solution 4

The fact that $a, b, c$ are the sides of a triangle doesn't do anything except telling you $b+c-a, c+a-b, a+b-c, a, b, c$ are all positive numbers (by triangle inequality). Once we established this, the problem become an ordinary inequality problem that will potentially have many ways to solve. Here is one solution:

We aim to prove:

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

(Here already we can finish the proof by Schur's inequality)

$\Rightarrow a^3+b^3+c^3+6abc \geq (a+b+c)(ab+bc+ca)$

This is homogeneous as both sides have the degree of 3. WLOG, we can thus let $a+b+c = 1$ , then the $RHS = ab+bc+ca$ .

The $LHS = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)+9abc = a^2+b^2+c^2-ab-bc-ca+9abc$ .

We need to prove $LHS \geq RHS \Rightarrow a^2+b^2+c^2-2ab-2bc-2ca+9abc = 1-4(ab+bc+ca)+9abc$

Using Lagrange multiplier, let $f(x) = 1-4(ab+bc+ca)+9abc+\lambda(a+b+c-1)$ , then we take the partial derivatives to obtain the four equations:

\begin{cases} \frac{\partial f}{\partial a} = -4(b+c)+9bc+\lambda = 0 \frac{\partial f}{\partial b} = -4(a+c)+9ac+\lambda = 0 \frac{\partial f}{\partial c} = -4(a+b)+9ab+\lambda = 0 \frac{\partial f}{\partial \lambda} = a+b+c-1 = 0 \end{cases}

Solving the system, we obtained $a=b=c=\frac{1}{3}$ or $a,b,c = \{5/18,5/18,4/9\}$ (Error compiling LaTeX. Unknown error_msg). Checking the boundaries we can also obtain $a,b,c = \{0,1/2,1/2\}$ (Error compiling LaTeX. Unknown error_msg), testing the values of $f(a,b,c)$ , we find that $f(a,b,c) \geq 0$ , and we are done.

We can verify that the interior stationary point $(1/3, 1/3, 1/3)$ is indeed a minimum by hessian matrix

~IDKHowtoaddsolution

Video Solution

https://youtu.be/6gDLBT1aGQM?si=ZR78mdotq4wfS3SA&t=637 [little fermat]

@@ 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==
-{{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 (Ravi Substitution) ==
+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.
+==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.
+==Solution 4==
+The fact that <math>a, b, c</math> are the sides of a triangle doesn't do anything except telling you <math>b+c-a, c+a-b, a+b-c, a, b, c</math> are all positive numbers (by triangle inequality). Once we established this, the problem become an ordinary inequality problem that will potentially have many ways to solve. Here is one solution:
+We aim to prove:
+<cmath>a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc} \Rightarrow a^3+b^3+c^3+3abc \geq a^2b+a^2c+b^2c+b^2a+c^2a+c^2b</cmath>
+(Here already we can finish the proof by Schur's inequality)
+<cmath>\Rightarrow a^3+b^3+c^3+6abc \geq (a+b+c)(ab+bc+ca)</cmath>
+This is homogeneous as both sides have the degree of 3. WLOG, we can thus let <math>a+b+c = 1</math>, then the <math>RHS = ab+bc+ca</math>.
+The <math>LHS = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)+9abc = a^2+b^2+c^2-ab-bc-ca+9abc</math>.
+We need to prove <cmath>LHS \geq RHS \Rightarrow a^2+b^2+c^2-2ab-2bc-2ca+9abc = 1-4(ab+bc+ca)+9abc</cmath>
+Using Lagrange multiplier, let <math>f(x) = 1-4(ab+bc+ca)+9abc+\lambda(a+b+c-1)</math>, then we take the partial derivatives to obtain the four equations:
+\begin{cases}
+\frac{\partial f}{\partial a} =        -4(b+c)+9bc+\lambda = 0
+\frac{\partial f}{\partial b} =        -4(a+c)+9ac+\lambda = 0
+\frac{\partial f}{\partial c} =        -4(a+b)+9ab+\lambda = 0
+\frac{\partial f}{\partial \lambda} = a+b+c-1 = 0
+\end{cases}
+Solving the system, we obtained <math>a=b=c=\frac{1}{3}</math> or <math>a,b,c = \{5/18,5/18,4/9\}</math>. Checking the boundaries we can also obtain <math>a,b,c = \{0,1/2,1/2\}</math>, testing the values of <math>f(a,b,c)</math>, we find that <math>f(a,b,c) \geq 0</math>, and we are done.
+*We can verify that the interior stationary point <math>(1/3, 1/3, 1/3)</math> is indeed a minimum by hessian matrix
+~[[User:IDKHowtoaddsolution|IDKHowtoaddsolution]]
+==Video Solution==
+https://youtu.be/6gDLBT1aGQM?si=ZR78mdotq4wfS3SA&t=637 [little fermat]
+== 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
All IMO Problems and Solutions

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