\(\left(2^a+3^a+4^a\right)\left(6^a+8^a+12^a\right)<24^{a+1}\)
\(\Leftrightarrow\left(2^a+3^a+4^a\right)\left(\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}\right)<24\)
Do \(a\in\left[1;2\right]\Rightarrow2\le2^a\le4;3\le3^a\le9;4\le4^a\le16\)
\(\Rightarrow2\le2^a\le16;2\le3^a\le16;2\le4^a\le16\)
Với \(x\in\left[2;16\right]\) ta có :
\(\left(x-2\right)\left(x-16\right)\le0\Leftrightarrow x^2-18x+32\le0\Leftrightarrow x-18+\frac{32}{x}\le0\Leftrightarrow\frac{32}{x}\le18-x\)
Từ đó suy ra :
\(32\left(\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}\right)<54-\left(2^a+3^a+4^a\right)\)
\(\Leftrightarrow\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}<\frac{54-\left(2^a+3^a+4^a\right)}{32}\)
\(\Leftrightarrow\left(2^a+3^a+4^a\right)\left(\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}\right)<\frac{\left(2^a+3^a+4^a\right)\left[54-\left(2^a+3^a+4^a\right)\right]}{32}\le\frac{1}{32}\left[\frac{\left(2^a+3^a+4^a\right)\left(54-\left(2^a+3^a+4^a\right)\right)}{2}\right]^2=\frac{729}{32}<24\)
