Can the Inequality x^x + y^y < (x+y)^(x+y) be Proven Algebraically?

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SUMMARY

The inequality x^x + y^y < (x+y)^(x+y) holds true for all x, y ≥ 1. Participants in the discussion suggest using derivatives to analyze the slope changes of both sides of the inequality. The expansion of (x+y)^(x+y) demonstrates that it is greater than x^x and y^y, as all terms in the expansion are positive when x and y are greater than 1. An algebraic proof is preferred over graphical methods for a definitive solution.

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bill01
Is it possible to prove this:
x^x + y^y < (x+y)^(x+y) for every x,y >=1 ?
 
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Well, let me think, since I am not really sure if this would work for all parameters of x and y... never mind, you have x,y >1! What you could do is take the derivative of both equations to measure it's change in slope, and if (x+y)^(x+y)is greater, then it will have a change in slope that is greater then the other equation. But I am not sure if that is what you want.
 
Iam just proving it,just expand the Right side (x+y)^(x+y), u get x multiplied by x+y times which is obviously greater than x^x and same in case of y and remaining terms of expansion are positive as x,y>1 and no negative terms in expansion. hope it helps.
 
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Thanks for the answers, but I would prefer an algebraic solution.
I did what Raul said with the graph but I would like an algebraic sol.
I believe that it is solved algebraically and it is not a transcendental equation.
 
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