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Difference between revisions of "1977 AHSME Problems/Problem 29"

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Anyways, look at which side is which. The squared side is smaller-- so that's good. It's in the right format.
 
Anyways, look at which side is which. The squared side is smaller-- so that's good. It's in the right format.
  
−
Cauchy's states that <math>(a_1b_1+a_2b_2+a_3b_3+......)^2 <= (a_1^2+a_2^2+a_3^2+....)(b_1^2+b_2^2+b_3^2+.....)</math>
+
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, 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>b1+b2+b3 = n</math>. So each of them is just 1, and <math>n = 3</math>-- answer choice B!
 
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>b1+b2+b3 = n</math>. So each of them is just 1, and <math>n = 3</math>-- answer choice B!

Revision as of 19:30, 30 January 2022

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

We see squares and one number. And we see an inequality. This calls for Cauchy's inequality. EEEEWWW.

Anyways, look at which side is which. The squared side is smaller-- so that's good. It's in the right format.

Cauchy's states that $(a_1b_1+a_2b_2+a_3b_3+......)^2 \le (a_1^2+a_2^2+a_3^2+....)(b_1^2+b_2^2+b_3^2+.....)$

Therefore, we see that, if we equate $a_1 = x^2, a_2 = y^2, a_3 = z^2$ 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 $b1+b2+b3 = n$. So each of them is just 1, and $n = 3$-- answer choice B!

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