Prove Schwarz Inequality for x, y, z in R+

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SUMMARY

The discussion focuses on proving the Schwarz Inequality for positive real numbers x, y, and z, specifically the inequality: √(x(3x+y)) + √(y(3y+z)) + √(z(3z+x)) ≤ 2(x+y+z). The participant initially expressed uncertainty about which inequality to apply and sought guidance on the standard approach. Ultimately, they successfully proved the inequality by utilizing the components of the Schwarz inequality and strategically assigning values to the variables involved.

PREREQUISITES
  • Understanding of the Schwarz Inequality in vector spaces
  • Familiarity with basic algebraic manipulation and inequalities
  • Knowledge of properties of positive real numbers
  • Experience with mathematical proof techniques
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  • Explore advanced proof techniques in inequality theory
  • Learn about vector spaces and their properties in relation to inequalities
  • Investigate other inequalities such as Cauchy-Schwarz and their proofs
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Homework Statement


For x,y,z ## \in \mathbb {R^+} ##, prove that
## \sqrt {x (3 x +y) } + \sqrt {y (3y +z) } + \sqrt {z(3z +x)} \leq ~ 2(x +y+ z) ##

Homework Equations


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The Attempt at a Solution


I don't know which inequality among the above two has to be applied.
I am trying to solve it by inspection. I don't know the standard approach to solve it.
 

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If in doubt, write the Schwarz inequality with components of general vectors v,w, expand the smaller side, then see if you can assign the components to get these square roots at a suitable spot in the equation.
 
mfb said:
If in doubt, write the Schwarz inequality with components of general vectors v,w, expand the smaller side, then see if you can assign the components to get these square roots at a suitable spot in the equation.

I did it. Thanks a lot for guiding me.
 

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