\left.\begin{matrix} \\ \end{matrix}\right\}

b^3 + c^3 - a^3 = a^2.b + a^2.c - a^3 &lt;=&gt; (b+c).(b^2 - bc + c^2 ) = a^2.(b+c) <=&gt; b^2 - bc+c^2 =a^2 &lt;=> bc= b^2 +c2 - a2 =&gt; bc = 2bc.cosA =&gt; cosA = 1/2 => A=60 độ (1)a = 2b.cosC => a= 2b .( a^2+b^2-c^2 )/2ab => a^2 = a^2+b^2-c^2 =&gt; b^2-c^2=0 =>b=c (2)Từ (1) và (2) => tam giác ABC đều

Áp dụng bđt (x+y+z).(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq9 , ta có (3-ab+3-bc+3-ca).(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\geq9 \Rightarrow(9-ab-bc-ca).(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\geq9 (1)Ta có a2+b2+c2\geqab+bc+ca\Rightarrowab+bc+ca\leq3 Từ (1)\Rightarrow(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\geq\frac{9}{9-ab-bc-ca}\geq\frac{3}{2}Dấu = xảy ra \Leftrightarrow a=b=c=1

Áp dụng bđt (x+y+z).(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 9 , ta có (3 - ab + 3 - bc + 3 - ca ).(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca} \geq 9 \Rightarrow (9 - ab - bc - ca ).(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca} \geq 9 (1)Ta có a2+b2+c2 \geq ab + bc + ca \Rightarrow ab+bc+ca \leq 3 Từ (1) \Rightarrow (\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca} \geq \frac{9}{9 - ab - bc - ca } \geq \frac{3}{2}Dấu = xảy ra \Leftrightarrow a=b=c=1

Áp dụng bđt (x+y+z).(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq9 , ta có (3-ab+3-bc+3-ca).(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\geq9 \Rightarrow(9-ab-bc-ca).(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\geq9 (1)Ta có a2+b2+c2\geqab+bc+ca\Rightarrowab+bc+ca\leq3 Từ (1)\Rightarrow(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\geq\frac{9}{9-ab-bc-ca}\geq\frac{3}{2}Dấu = xảy ra \Leftrightarrow a=b=c=1