(a) The inequality follows from the binomial theorem.
[tex](a - b)^2 = a^2 - 2ab + b^2 \ge 0 \implies a^2 + 2ab + b^2 = (a+b)^2 \ge 4ab \\\\ \implies \dfrac{(a+b)^2}4 \ge ab \\\\ \implies \dfrac{a+b}2 \ge \sqrt{ab} ~~~~~~~~ (1)[/tex]
By the same reasoning,
[tex]\dfrac{c+d}2 \ge \sqrt{cd}[/tex]
Adding these results together, we have
[tex]\dfrac{a+b}2 + \dfrac{c+d}2 = \dfrac{a+b+c+d}2 \ge \sqrt{ab} + \sqrt{cd}[/tex]
and since [tex]\sqrt{ab}[/tex] and [tex]\sqrt{cd}[/tex] are both positive, they also satisfy the AM-GM inequality,
[tex]\dfrac{\sqrt{ab} + \sqrt{cd}}2 \ge \sqrt{\sqrt{ab}\times\sqrt{cd}} = \sqrt[4]{abcd}[/tex]
and the 4-variable result follows,
[tex]\dfrac{a+b+c+d}2 \ge 2\times\dfrac{\sqrt{ab}+\sqrt{cd}}2 \ge 2 \sqrt[4]{abcd} \\\\ \implies \dfrac{a+b+c+d}4 \ge \sqrt[4]{abcd} ~~~~~~~~ (2)[/tex]
(b.i) With [tex]m=\frac{a+b+c}3[/tex], we have
[tex]\dfrac{3m}4 = \dfrac{a+b+c}4 \implies m - \dfrac m4 = \dfrac{a+b+c}4 \\\\ \implies m = \dfrac{a+b+c+m}4[/tex]
(b.ii) From (2) it follows that
[tex]\dfrac{a+b+c+m}4 \ge \sqrt[4]{abcm} = \sqrt[4]{abc\times\dfrac{a+b+c+m}4}[/tex]
Using the result from (b.i), this is equivalent to
[tex]\dfrac{a+b+c}3 \ge \sqrt[4]{abc\times\dfrac{a+b+c}3}[/tex]
Take 4th powers on both sides.
[tex]\left(\dfrac{a+b+c}3\right)^4 \ge abc \times \dfrac{a+b+c}3[/tex]
Divide both sides by [tex]\frac{a+b+c}3[/tex].
[tex]\left(\dfrac{a+b+c}3\right)^3 \ge abc[/tex]
Finally, take the cube root of both sides.
[tex]\dfrac{a+b+c}3 \ge \sqrt[3]{abc}[/tex]
as required.
