Prove that $a^{\lambda b} + b^{\lambda a} + a^{\lambda b^2} + b^{\lambda a^2} \le 2$ for positive reals $a+b=1$ 
Problem 1: Let $\lambda = \frac{\ln (7 + 3\sqrt{5})}{\ln 2} - 1 \approx 2.77697$. Let $a, b$ be positive real numbers with $a + b = 1$. Prove (or disprove) that
$$a^{\lambda b} + b^{\lambda a} + a^{\lambda b^2} + b^{\lambda a^2} \le 2$$
with equality if and only if $a = b = \frac{1}{2}$


Problem 2(a weaker version of Problem 1): Let $\lambda_1 = \frac{25}{9}$. Let $a, b$ be positive real numbers with $a + b = 1$. Prove (or disprove) that
$$a^{\lambda_1 b} + b^{\lambda_1 a} + a^{\lambda_1 b^2} + b^{\lambda_1 a^2} \le 2.$$

Background Information: In Proposition 5.2 in [1], Vasile Cirtoaje gives the following result:
Problem 3: If $a, b$ are nonnegative real numbers satisfying $a + b = 1$, then $a^{2b} + b^{2a} \le 1$.
In Conjecture 5.1 in [1], Vasile Cirtoaje proposes the following conjecture:
Let $a, b$ be nonnegative real numbers satisfying $a + b = 1$. If $k \ge 1$, then $a^{(2b)^k} + b^{(2a)^k} \le 1$.
The case $k=2$ has been proved:
Problem 4: Let $a, b$ be positive real numbers with $a + b = 1$. Prove that $a^{4b^2} + b^{4a^2} \le 1$.
See If $a+b=1$ so $a^{4b^2}+b^{4a^2}\leq1$
I combine Proposition 5.2 and Problem 4 to come up with the problems.
I may use appropriate bounds to prove Problems 3 and 4 (something like
Inequality $a^{2b}+b^{2a}\leq \cos(ab)^{(a-b)^2}$). Now the problems are more difficult.
Reference
[1] Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions",
The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137.
https://eudml.org/doc/223938

working the weaker case and using $b=1-a$, we have
$$f(a)=a^{\frac{25}{9} (1-a)}+(1-a)^{\frac{25}{9}a  }+a^{\frac{25}{9} (1-a)^2}+(1-a)^{\frac{25 }{9}a^2}$$
The derivative cancels close to $a=\frac 14$. In fact, the first iterate of Newton method is extremely close to $\frac 6 {25}$
$$f\left(\frac{1}{4}\right)\sim 1.94050 \qquad \text{and} \qquad 
f\left(\frac6{25}\right)\sim 1.94029$$
