# What are all real functions that show that f(x+y)=<f(xy)?please,give exact answer

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Plugging y = 0 in the relation yields: `f(x)<= f(0)`

Replacing y by `-sqrtx ` and x by `sqrt x` yields: `f(sqrtx - sqrtx) = f(0)<= f(-x)=gt f(-x) = 0.`

Replacing y by `-sqrtx` and x by `-sqrt x` yields: `f(0) = f(sqrtx+sqrtx) = f(0)<= f((-sqrtx)*(-sqrtx)) = f(x) =gt f(x) = 0` .

If f(x) = f(0) => the function is constant.

**The conclusion to this study is that only the constant functions, f(x) = a, check the relation `f(x+y) <=f(xy).` **