70th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

We will show that the maximum value is $\frac{1}{16}$, obtained when $a=b=\frac{1}{2}$.

The inequality $E(a,b)\le \frac{1}{16}$ is equivalent to $16(a+b)\le (4a^2+3)(4b^2+3)$, which we rewrite as $(4ab-1{)}^2+4(a+b-1{)}^2+2(2a-1{)}^2+2(2b-1{)}^2\ge 0$, obviously true.

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Alternative solution.

We will show that the maximum value is $\frac{1}{16}$, obtained when $a=b=\frac{1}{2}$.

We combine the following inequalities: $a+b\le \frac{(1+a+b{)}^2}{4}$ and (from CBS) $(4a^2+3)(4b^2+3)=(4a^2+1+2)(1+4b^2+2)\ge (2a+2b+2{)}^2$. We get: $$\frac{a+b}{(4a^2+3)(4b^2+3)}\le \frac{(a+b+1{)}^2}{4(4a^2+3)(4b^2+3)}\le \frac{(a+b+1{)}^2}{4(2a+2b+2{)}^2}\le \frac{1}{16}.$$