Câu 5.

Ta có: $P=(\sqrt{a^2+abc}+\sqrt{abc})+(\sqrt{b^2+abc}+\sqrt{abc})+(\sqrt{c^2+abc}+\sqrt{abc})+6\sqrt{abc}$
Áp dụng BĐT Cauchy ta được:
$\sqrt{abc}\le \sqrt{\left(\dfrac{a+b+c}{3}\right)^3}=\sqrt{\dfrac{1}{27}}=\dfrac{1}{3\sqrt{3}}\\\Rightarrow 6\sqrt{abc}\le \dfrac{2.3}{3\sqrt{3}}=\dfrac{2}{\sqrt{3}}$
Ta có: $\sqrt{a^2+abc}+\sqrt{abc}$
$=\sqrt{a}(\sqrt{a+bc}+\sqrt{bc})\\=\sqrt{a}(\sqrt{(a+b)(a+c)}+\sqrt{bc})$
Áp dụng BĐT Cauchy ta được:
$\sqrt{a.\dfrac{1}{3}}\le \dfrac{1}{2}\left(\dfrac{1}{3}+a\right)\\\Rightarrow \sqrt{a}\le \dfrac{\sqrt{3}}{2}\left(\dfrac{1}{3}+a\right)$
$\sqrt{(a+b)(a+c)}\le \dfrac{a+b+a+c}{2}=\dfrac{2a+b+c}{2}\\\sqrt{bc}\le \dfrac{b+c}{2}$
Do đó: $\sqrt{a^2+abc}+\sqrt{abc}\le \dfrac{2a+2b+2c}{2}=a+b+c=1$
$\Rightarrow \sqrt{a^2+abc}+\sqrt{abc}\le \dfrac{\sqrt{3}}{2}\left(\dfrac{1}{3}+a\right)$
Tương tự và cộng vế:
$P\le \dfrac{\sqrt{3}}{2}\left(\dfrac{1}{3}.3+a+b+c\right)+\dfrac{2}{\sqrt{3}}=\dfrac{\sqrt{3}}{2}(1+1)+\dfrac{2}{\sqrt{3}}=\dfrac{5\sqrt{3}}{3}$
Dấu "=" có khi $a=b=c=\dfrac{1}{3}$
Vậy $P_{max}=\dfrac{5\sqrt{3}}{3}$ khi $a=b=c=\dfrac{1}{3}$