Ta có $\bullet\;\sum \frac{a^2}{b+c} -\frac{a+b+c}{2}$$=\sum \Bigg(\frac{a^2}{b+c}-\frac{b+c-4a}{4} \Bigg)= \sum \frac{(2a-b-c)^2}{4(b+c)}$$\bullet\;\frac{\sqrt{3(a^2+b^2+c^2)}}{2}-\frac{a+b+c}{2}= \frac{3(a^2+b^2+c^2)-(a+b+c)^2}{2\Big[\sqrt{3(a^2+b^2+c^2)}+(a+b+c) \Big]}$$\le \frac{(2a-b-c)^2+(2b-c-a)^2+(2c-a-b)^2}{12(a+b+c) } $Nên chỉ cần cm$ \sum \frac{(2a-b-c)^2}{4(b+c)} \ge \sum \frac{(2a-b-c)^2}{12(a+b+c) }$$\Leftrightarrow \sum(2a-b-c)^2\left(\frac{1}{b+c}- \frac{1}{3(a+b+c)} \right) \ge0$BDT cuối luôn đúng, đẳng thức xảy ra $\Leftrightarrow a=b=c=1$
Ta có $\bullet\;\sum \frac{a^2}{b+c} -\frac{a+b+c}{2}$$=\sum \Bigg(\frac{a^2}{b+c}-\frac{b+c-4a}{4} \Bigg)= \sum \frac{(2a-b-c)^2}{4(b+c)}$$\bullet\;\frac{\sqrt{3(a^2+b^2+c^2)}}{2}-\frac{a+b+c}{2}= \frac{3(a^2+b^2+c^2)-(a+b+c)^2}{2\Big[\sqrt{3(a^2+b^2+c^2)}+(a+b+c) \Big]}$$=\frac{(2a-b-c)^2+(2b-c-a)^2+(2c-a-b)^2}{6\Big[\sqrt{3(a^2+b^2+c^2)}+(a+b+c) \Big]}$$\therefore\sum \frac{a^2}{b+c} \ge \frac{\sqrt{a^2+^2+c^2}}{2}$$\Longleftrightarrow \sum \frac{(2a-b-c)^2}{4(b+c)} \ge \sum \frac{(2a-b-c)^2}{6\Big[\sqrt{3(a^2+b^2+c^2)}+(a+b+c) \Big]}$$\Leftrightarrow \sum(2a-b-c)^2\left(\frac{1}{2(b+c)}- \frac{1}{3\left[\sqrt{3(a^2+b^2+c^2)} +(a+b+c)\right]} \right) \ge0$BDT cuối luôn đúng, đẳng thức xảy ra $\Leftrightarrow a=b=c=1$
Ta có $\bullet\;\sum \frac{a^2}{b+c} -\frac{a+b+c}{2}$$=\sum \Bigg(\frac{a^2}{b+c}-\frac{b+c-4a}{4} \Bigg)= \sum \frac{(2a-b-c)^2}{4(b+c)}$$\bullet\;\frac{\sqrt{3(a^2+b^2+c^2)}}{2}-\frac{a+b+c}{2}= \frac{3(a^2+b^2+c^2)-(a+b+c)^2}{2\Big[\sqrt{3(a^2+b^2+c^2)}+(a+b+c) \Big]}$$
\le \frac{(2a-b-c)^2+(2b-c-a)^2+(2c-a-b)^2}{
12(a+b+c) }
$
Nên c
hỉ c
ần c
m$ \sum \frac{(2a-b-c)^2}{4(b+c)} \ge \sum \frac{(2a-b-c)^2}{
12(a+b+c) }$$\Leftrightarrow \sum(2a-b-c)^2\left(\frac{1}{b+c}- \frac{1}{3(a+b+c)} \right) \ge0$BDT cuối luôn đúng, đẳng thức xảy ra $\Leftrightarrow a=b=c=1$