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1977 AHSME Problems/Problem 29: Difference between revisions

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[[1977 AHSME Problems/Problem 29|Solution]]
[[1977 AHSME Problems/Problem 29|Solution]]


==Solution==
==Solution (Official MAA)==


We see squares and one number. And we see an inequality. This calls for Cauchy's inequality. EEEEWWW.
Let <math>a = x^2</math>, <math>b = y^2</math>, <math>c = z^2</math>. Then


Anyways, look at which side is which. The squared side is smaller-- so that's good. It's in the right format.
<cmath> 0 \leq (a-b)^2 + (b-c)^2 + (c-a)^2</cmath>
<cmath>\dfrac{ab+bc+ca}{a^2 + b^2 + c^2} \leq 1;</cmath>
<cmath>\dfrac{a^2 + b^2 + c^2 + 2(ab+bc+ca)}{a^2 + b^2 + c^2} \leq 3;</cmath>
<cmath>(a+b+c)^2 \leq 3(a^2 + b^2 + c^2).</cmath>


Cauchy's states that <math>(a_1b_1+a_2b_2+a_3b_3+......)^2 \le (a_1^2+a_2^2+a_3^2+....)(b_1^2+b_2^2+b_3^2+.....)</math>
Therefore <math>n\leq 3</math>. Choosing <math>a = b = c > 0</math> shows <math>n</math> is not less than three.
 
Therefore, we see that, if we equate <math>a_1 = x^2, a_2 = y^2, a_3 = z^2</math> we get the equality right away. What's the final step? Figuring out this n. Now, note that the equation is basically complete; all we need is for <math>b_1+b_2+b_3 = n</math>. So each of them is just 1, and <math>n = 3</math>-- answer choice <math>\boxed{B}</math>!

Latest revision as of 11:21, 21 April 2024

Problem 29

Find the smallest integer $n$ such that $(x^2+y^2+z^2)^2\le n(x^4+y^4+z^4)$ for all real numbers $x,y$, and $z$.

$\textbf{(A) }2\qquad \textbf{(B) }3\qquad \textbf{(C) }4\qquad \textbf{(D) }6\qquad \textbf{(E) }\text{There is no such integer n}$

Solution

Solution (Official MAA)

Let $a = x^2$, $b = y^2$, $c = z^2$. Then

\[0 \leq (a-b)^2 + (b-c)^2 + (c-a)^2\] \[\dfrac{ab+bc+ca}{a^2 + b^2 + c^2} \leq 1;\] \[\dfrac{a^2 + b^2 + c^2 + 2(ab+bc+ca)}{a^2 + b^2 + c^2} \leq 3;\] \[(a+b+c)^2 \leq 3(a^2 + b^2 + c^2).\]

Therefore $n\leq 3$. Choosing $a = b = c > 0$ shows $n$ is not less than three.

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