Black Franklin
05/12/2023 · Junior High School
\( \sqrt { | x y | } \leq \frac { 1 } { \sqrt { 2 } } \sqrt { x ^ { 2 } + y ^ { 2 } } \)
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Tutor-Verified Answer
Step-by-step Solution
To prove the inequality
\[
\sqrt{|xy|} \leq \frac{1}{\sqrt{2}} \sqrt{x^2 + y^2},
\]
we can start by squaring both sides to eliminate the square roots. This gives us:
\[
|xy| \leq \frac{1}{2} (x^2 + y^2).
\]
Next, we can rearrange this inequality:
\[
2|xy| \leq x^2 + y^2.
\]
Now, we can apply the AM-GM (Arithmetic Mean-Geometric Mean) inequality, which states that for any non-negative real numbers \(a\) and \(b\):
\[
\frac{a + b}{2} \geq \sqrt{ab}.
\]
In our case, let \(a = x^2\) and \(b = y^2\). Then, we have:
\[
\frac{x^2 + y^2}{2} \geq \sqrt{x^2y^2} = |xy|.
\]
Multiplying both sides by 2 gives:
\[
x^2 + y^2 \geq 2|xy|.
\]
This shows that:
\[
2|xy| \leq x^2 + y^2,
\]
which is exactly what we needed to prove. Therefore, we conclude that:
\[
\sqrt{|xy|} \leq \frac{1}{\sqrt{2}} \sqrt{x^2 + y^2}.
\]
Thus, the inequality holds true.
Quick Answer
The inequality \( \sqrt{|xy|} \leq \frac{1}{\sqrt{2}} \sqrt{x^2 + y^2} \) holds true.
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