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1)
Show that [5x]+ [5y] is greater than or equal to [3x+y] + [3y+x] where x,y greater than or equal to 0..
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12 Answers
1057arre dis is easy...
[5x]+[5y] ≤ 5(x+y)
[3x+y]+[3y+x] ≤ 4(x+y)
and obviusly 5(x+y) is greater than 4(x+y) for all x,y≥0
thus the result follows...
the equality holds when x=y=0
21Ketan, please tell me why if 5(x+y)≥4(x+y) then the rest follows?
1057@arnab...[5x]+[5y] ≤ 5(x+y) and [3x+y]+[3y+x] ≤ 4(x+y)
5(x+y)≥4(x+y)
thus [5x]+[5y]≥[3x+y]+[3y+x] from the first two inequalities....
1708\hspace{-16}\mathbf{x-1<[x]\leq x}\forall x\in\mathbb{R}$\\\\ So $\mathbf{5x-1<[5x]}$ and $\mathbf{5y-1<[5y]}$\\\\ So $\mathbf{5(x+y)-2<[5x]+[5y]}$\\\\ OR $\mathbf{[5x]+[5y]>5(x+y)-2}$\\\\ Similarly \\\\ $\mathbf{[3x+y]<3x+y}$\\\\ $\mathbf{[3y+x]<3y+x}$\\\\ So $\mathbf{[3x+y]+[3y+x]<4(x+y)}$\\\\ Now Given $\mathbf{[5x]+[5y]\geq [3x+y]+[3y+x]}$\\\\ So $\mathbf{5(x+y)-2\geq 4(x+y)}$\\\\ $\mathbf{x+y-2\geq 0}$\\\\ Now We have to show that this Inequality is true $\mathbf{\forall x,y\in \left[0,\infty)}$
71Within the given domain, how will this inequality hold? Only possible if x,y ≥ 1 . Isn't it?
1708Yes Vivek you are saying Right.
But i think the proof is still not Complete.
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