Proving $(a+b)\log_2(a+b) \ge a\log_2 a +b\log_2 b +2a $ for positive integers $b \ge a$ On the 6th page of the paper &quot;On Induced Subgraphs of the Cube&quot; (PDF link via usdc.edu) (precisely: the last line of Lemma 4.1) the author uses an inequality:
$$(a+b)\log_2(a+b) \ge a\log_2 a +b\log_2 b +2a $$ where $b \ge a$ is known.
I'm not sure how that holds. Can someone explain this calculation to me?
Additional information: $a$ and $b$ are number of vertices of graph (respectively $|V_1|$ and $|V_2|$ in the paper) so they are both positive integers.
With the substitution $b=ka$, $k \ge 1$, the inequality becomes
$$
 (k+1)a \left(\log_2 (k+1) + \log_2 a \right) \ge a \log_2 a + k a \left( \log_2 k + \log_2 a \right) + 2a
$$
which reduces to
$$
 (k+1) \log_2(k+1) \ge k \log_2 k + 2
$$
or
$$
 \frac{(k+1)^{k+1}}{k^k} \ge 4 \, .
$$
The last inequality is true for $k\ge 1$ because
$$
\frac{(k+1)^{k+1}}{k^k} = (k+1) \left( 1 + \frac 1k \right)^k \ge 2 \cdot 2 = 4 
$$
(using the binomial formula or Bernoulli's inequality for the second factor).
The fact that $a$ and $b$ are integers is not needed for the estimate, only that $b \ge a &gt; 0$.
This is  true for naturals (The cited paper is on graph theory.), as the result of the command of Mathematica
NMinimize[{(a + b)*Log[2, a + b] - a*Log[2, a] - b*Log[2, b] - 2 a, 
b &gt;= a &amp;&amp;{a, b} \[Element] PositiveIntegers}}, {a, b}]

$\{3.552713678800501\cdot 10^{-15},\{a\to 4,b\to 4\}\}$
shows.