Giải câu 5 đề 5 ôn thi toán lớp 9 lên 10

Ta có: $(a - b)^{2} ≥ 0 => a^{2} + b^{2} ≥ 2ab$

$a^{3} + b^{3} + abc = (a + b)(a^{2} - ab + b^{2} ) + abc$

$a^{3} + b^{3} + abc ≥ (a + b)(2ab - ab) + abc = ab(a + b) + abc$

=>$ a^{3} + b^{3} + abc ≥ ab(a + b + c)$

Vì a, b, c > 0 nên:

$\frac{1}{a^{3}+b^{3}+abc}\leq \frac{1}{ab(a+b+c)}$

Tương tự, ta có:

$\frac{1}{b^{3}+c^{3}+abc}\leq \frac{1}{bc(a+b+c)}$

$\frac{1}{c^{3}+a^{3}+abc}\leq \frac{1}{ca(a+b+c)}$

=> $\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}abc}+\frac{1}{c^{3}+a^{3}+abc}\leq \frac{1}{abc}$

$\leq \frac{1}{ab(a+b+c)}+\frac{1}{bc(a+b+c)}+\frac{1}{ca(a+b+c)}\leq \frac{a+b+c}{abc(a+b+c)}=\frac{1}{abc}$

Dấu bằng xảy ra khi a = b = c