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 | ||
Line 17: | Line 35: | ||
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. | ||
− | ==Solution | + | eevee9406: This solution uses incorrect reasoning regarding the inequalities. |
+ | |||
+ | ==Solution 3== | ||
Rearrange to get | Rearrange to get | ||
<cmath>a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \ge 0,</cmath> | <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. | which is true by Schur's inequality. | ||
− | == See Also == {{IMO box|year= | + | == See Also == {{IMO box|year=1964|num-b=1|num-a=3}} |
Latest revision as of 17:32, 4 July 2024
Problem
Suppose are the sides of a triangle. Prove that
Solution
Let ,
, and
. Then,
,
, and
. By AM-GM,
Multiplying these equations, we have
We can now simplify:
~mathboy100
Solution 2
We can use the substitution ,
, and
to get
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
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 |