False or true proof of $x^{2(1-x)}+(1-x)^{2(x)}<1$

Question

Sketch of proof : As the function (see above) is symetric we consider only $0< x < 0.5$ . We start with two lemma

Lemma 1 : $$x^{2(1-x)}<x$$

Lemma 2 : $$(1-x)^{2(x)}>1-x$$

this two lemma are not hard to show omited .

So we have $x^{2(1-x)}\neq x$ and $(1-x)^{2(x)}\neq 1-x$ it implies $x^{2(1-x)}+(1-x)^{2(x)}\neq 1-x+x=1$

As the function $x^{2(1-x)}+(1-x)^{2(x)}=f(x)$ is continuous it remains to calculate a value of $f(x)$ to end the proof .

Is it right ?

Thanks in advance for your time and patience .

Answer 1

If I evaluate the left expression at 3, I get $64+{{1}\over{81}}$

So maybe if you restrict it to $0\le{x}\le1$. Note that outside that range, the result isn't real unless $2x$ is an integer.

If I evaluate the left expression at 0, 0.5, and 1, I get 1.

So even restricted to $0\lt{x}\lt1$, I would say false.

Answer 2

Your reasoning on inequalities is wrong.

$$4\ne3\land6\ne7$$ but $$4+6=3+7.$$