I am confused by the meaning of the $t \in (0,1)$ parameter in Jensen's inequality $$ f( tx_1+(1-t)x_2) \le tf(x_1)+(1-t)f(x_2) $$ When I apply this to the logarithm $$ \log( tx_1+(1-t)x_2) \ge t \log(x_1)+(1-t) \log(x_2).$$ Does this mean the convex version of the inequality does not hold for $$ \log( x_1+x_2) \le \log(x_1)+ \log(x_2)?$$ Are there any results for the relation of the quantities on the left and the right si de of the last inequality?
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$\begingroup$The inequality you want to have $$\log(x_1+x_2)\leq \log(x_1)+\log(x_2)$$ is equivalent to $$x_1+x_2\leq x_1x_2$$ so it holds whenever the latter one holds, for example when $x_1,x_2\geq 2$. Jensen has nothing to do with this.
$\endgroup$ $\begingroup$The correct inequality is $$ \log( tx_1+(1-t)x_2) \ge t \log(x_1)+(1-t) \log(x_2).$$In particular we have $$ \log( \frac {x_1+x_2} 2) \ge \frac 1 2 (\log(x_1)+ \log(x_2)).$$
Your last in equality fails when $x_1=x_2=\frac 1 2$.
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