Can var(x+y) Be Less Than or Equal to 2(var(x) + var(y))?

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The discussion focuses on proving the inequality var(x+y) ≤ 2(var(x) + var(y)). The proof begins with the equation var(x+y) = var(x) + var(y) + 2cov(x,y) and utilizes the Cauchy-Schwarz inequality to establish that cov(x,y) squared is less than or equal to the product of var(x) and var(y). By manipulating the terms, it is shown that 2cov(x,y) can be bounded by the sum of the variances, leading to the desired inequality. The conclusion reinforces the relationship between variances and covariance, highlighting the importance of the Cauchy-Schwarz inequality in this context. This proof effectively demonstrates the stated inequality.
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Hi, I was hoping that someone might be able to please help me with this proof.

Prove that var(x+y) ≤ 2(var(x) + var(y)).

So far I have:

var(x+y) = var(x) + var(y) + 2cov(x,y)

where the cov(x,y) = E(xy) - E(x)E(y), but I'm not really sure to go from there.
Any insight would be very helpful!

Thanks!
 
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Let \alpha=var(x),\beta=var(y),\gamma=cov(x,y). According to the Cauchy-Schwarz inequality (see http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality for the proof), \gamma^{2}\leq{}\alpha\beta. We want to show \alpha{}+\beta{}+2\gamma\leq{}2(\alpha{}+\beta), which follows directly from

2\gamma\leq{}2(\gamma^{2})^{\frac{1}{2}}\leq{}2(\alpha\beta)^{\frac{1}{2}}\leq{}2(\frac{\alpha{}+\beta}{2})\leq{}\alpha{}+\beta

where (\alpha\beta)^{\frac{1}{2}}\leq{}\frac{\alpha{}+\beta}{2} follows from the well-known fact that the geometric mean is always smaller than the arithmetic mean (see http://www.cut-the-knot.org/pythagoras/corollary.shtml for proof).
 
Awesome! Thanks so much for your help!
 
You are welcome.
 
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