Power of arithmetic mean Does there exist a finite universal constant $m>1$ such that for every $0\leq a,\lambda\leq 1$ the following holds? $$[a\lambda+(1-a)(1-\lambda)]^m\leq \lambda^a(1-\lambda)^{1-a}$$
My guess is that perhaps there exists a small $m$ since $a\lambda+(1-a)(1-\lambda)\leq 1$ since as $m$ increases the power of the arithmetic mean decreases. 
I have tried to apply Jensen's and Holder's inequality but nothing much can be said. 
 A: This answers the original version of the question.
Graphical view
$\require{extpfeil}\Newextarrow{\xlr}{5,5}{0x2194}$
$$\star \triangleq a\lambda+(1-a)(1-\lambda)$$
\begin{array}{ccccccccc}
0 & — & \lambda & \xlr{1-a} & \star & \xlr{a} & 1-\lambda & — & \rlap{1 \to x\text{-axis}}\\
| & & | & & | & & \downarrow & \color{red}{\nearrow} \\
| & & | & & | & & \color{red}{\log(1-\lambda)} \\
| & & | & & \downarrow & \color{red}{\nearrow} \color{blue}{/} \\
| & & | & & \llap{\color{red}{\log\star}} & \color{blue}{/} \\
| & & | & \quad\color{red}{/} & \downarrow & \!\!\!\!\!\color{blue}{/} \\
| & & | & \color{red}{/} & \color{blue}{\bullet \; \rlap{\xleftarrow{\text{weight mean}} a\log\lambda+(1-a)\log(1-\lambda)}} \\
| & & \downarrow & \color{red}{/} \color{blue}{\nearrow} & \\
| & & \color{red}{\log\lambda} & & & \rlap{\text{Legend}} \\
\downarrow & \color{red}{\swarrow} & & & \rlap{\color{red}{\text{concave function }f(x) = \log(x)}} \\
\llap{\text{negative $y$-axis}} & & & & \rlap{\color{blue}{\text{straight line joining }} \color{red}{\log\lambda \text{ and } \log(1-\lambda)}}
\end{array}
From the above "graph" of the concave function $\log$, it's observed that
$$0 \ge \color{red}{\underbrace{\log[a\lambda+(1-a)(1-\lambda)]}_{\Large\log\star}} \ge \color{blue}{\underbrace{a\log\lambda+(1-a)\log(1-\lambda)}_{\Large \bullet \text{ weighted mean}}}, \tag1 \label1$$
which is an application Jensen's inequality on the concave function $\log$.
Take $\log$ on the original problem
\begin{align}
[a\lambda+(1-a)(1-\lambda)]^m &\le \lambda^a(1-\lambda)^{1-a} \\
\iff m\, \color{red}{\underbrace{\log[a\lambda+(1-a)(1-\lambda)]}_{\Large\log\star}} &\le \color{blue}{\underbrace{a\log\lambda+(1-a)\log(1-\lambda)}_{\Large \bullet \text{ weighted mean}}} \le 0 \tag2
\end{align}
To answer the question, just take any $m \ne 1$ so that $$m \ge \frac{a\log\lambda+(1-a)\log(1-\lambda)}{\log[a\lambda+(1-a)(1-\lambda)]} \stackrel{\eqref{1}}{\ge} 1 \tag3$$
as long as $\color{red}{\log\star} = \log[a\lambda+(1-a)(1-\lambda)]$ is defined and nonzero.
Two extreme cases when the above argument fails


*

*$\star = 0$, so $\color{red}{\log\star}$ is undefined.  In this case, $a\lambda = (1-a)(1-\lambda)=0$  Either $(a,\lambda)=(1,0)$ or $(a,\lambda)=(0,1)$, so $\lambda(1-\lambda)=0$, and the desired inequality holds for any $m > 0$.

*$\star = 1$, so $\color{red}{\log\star} = 0$
\begin{align}
1 &= a+(1-a) \tag4 \\
&= a\lambda + (1-a)(1-\lambda) \tag5 \\
(4)-(5): 0 &= a(1-\lambda)+(1-a)\lambda \\
\therefore \begin{cases} a=0\\\lambda=0 \end{cases} &\text{ or } \begin{cases} a=1\\\lambda=1 \end{cases}
\end{align}
In this case, the LHS of the desired inequality equals one for any $m \in \Bbb R$, while the RHS, containing $0^0$, becomes undefined.
