Let $a,b,t>0$. If $0\leq\lambda\leq 1$, then $a^\lambda b^{1-\lambda}\leq\lambda t^{1/\lambda}a+(1-\lambda)t^{-1/(1-\lambda)}b$ where equality holds for $t^{1/\lambda}a=t^{-1/(1-\lambda)} b$.
$\textbf{Q:}$ First, the inequality seems not making a lot of sense when $\lambda=0$ or $\lambda=1$. My assumption is that $0<\lambda<1$. With $t=1$, it is easy to see the inequality by convexity argument. I am considering $(t^{1/\lambda}a)^\lambda(t^{-1/(1-\lambda)}b)^{1/(1-\lambda)}$ for convexity argument but evaluation at the end point does not make any sense here. The book says this is given by comparison of arithmetic and geometric means. How should I derive above inequality?
Ref. Kodaira Complex Manifolds and Deformation of Complex Structures, Appendix Sec 1, Lemma 1.1 on pg 377.
You are close. Write $$ a^\lambda b^{1-\lambda} = (t^{1/\lambda} a)^\lambda \cdot (t^{-1/(1-\lambda)} b)^{1-\lambda} $$ and then use the concavity of the logarithm: $$ \log \left(a^\lambda b^{1-\lambda}\right) = \lambda \log\left(t^{1/\lambda} a\right) + (1-\lambda) \log\left(t^{-1/(1-\lambda)} b\right) \le \log \left(\lambda t^{1/\lambda} a + (1-\lambda)t^{-1/(1-\lambda)} b \right) $$