Chủ đề này có 2 trả lời

[leduylinh1998]

Cho a, b,c dương thoả mãn: $\sum \sqrt{a^{2}+b^{2}}=\sqrt{2011}$. CMR:$\sum \frac{a^{2}}{b+c}\geq \frac{1}{2}\sqrt{\frac{2011}{2}}$

[Hoang Tung 126]

Theo bđt Bunhiacopxki ta có :$\sum \frac{a^2}{b+c}\geq \sum \frac{a^2}{\sqrt{2(b^2+c^2)}}=\sum \frac{a^4}{a.\sqrt{2(a^2b^2+a^2c^2)}}\geq \frac{(a^2+b^2+c^2)^2}{\sum a.\sqrt{2(a^2b^2+a^2c^2)}}$

Mặt khác :$\sum a\sqrt{2(a^2b^2+a^2c^2)}\leq \sqrt{(a^2+b^2+c^2).2(2a^2b^2+2b^2c^2+2c^2a^2)}=\sqrt{4(a^2+b^2+c^2).(a^2b^2+b^2c^2+c^2a^2)}\leq \sqrt{4(a^2+b^2+c^2).\frac{(a^2+b^2+c^2)^2}{3}}=\sqrt{\frac{4}{3}.(a^2+b^2+c^2)^3}= > \frac{(a^2+b^2+c^2)^2}{\sum a.\sqrt{2(a^2b^2+a^2c^2)}}\geq \frac{(a^2+b^2+c^2)^2}{\sqrt{\frac{4}{3}.(a^2+b^2+c^2)^3}}=\frac{\sqrt{(a^2+b^2+c^2)}}{\sqrt{\frac{4}{3}}}=\frac{\sqrt{3(a^2+b^2+c^2)}}{2}$(1)

Mà :$\sqrt{2011}=\sum \sqrt{a^2+b^2}\leq \sqrt{3(2a^2+2b^2+2c^2)}=\sqrt{6(a^2+b^2+c^2)}= > a^2+b^2+c^2\geq \frac{2011}{6}$(2)

Từ (1),(2)$= > \sum \frac{a^2}{b+c}\geq \frac{1}{2}.\sqrt{\frac{2011}{2}}$

Dấu = xảy ra khi a=b=c

[leduylinh1998]

Đặt: $\sqrt{a^{2}+b^{2}}=x;\sqrt{b^{2}+c^{2}}=y;\sqrt{c^{2}+a^{2}}=z$    ( x,y,z>0)

$\Rightarrow x+y+z=\sqrt{2011}$

$\Rightarrow \left\{\begin{matrix} a^{2}=\frac{z^{2}+x^{2}-y^{2}}{2} & \\ b^{2}=\frac{x^{2}+y^{2}-z^{2}}{2}& \\ c^{2}=\frac{y^{2}+z^{2}-x^{2}}{2}& \end{matrix}\right.$

Dễ chứng minh được: $\sqrt{2(m^{2}+n^{2})}\geq m+n$

$\Rightarrow \sum \frac{a^{2}}{b+c}\geq \sum \frac{z^{2}+x^{2}-y^{2}}{2\sqrt{2}.y}\geq \frac{1}{2\sqrt{2}}\sum \frac{x^{2}+y^{2}+z^{2}}{x+y+z}\geq \frac{1}{2\sqrt{2}}\sum \frac{(x+y+z)^{2}}{x+y+z}\geq \frac{1}{2\sqrt{2}}(x+y+z)=\frac{1}{2\sqrt{2}}.\sqrt{2011}=\frac{1}{2}\sqrt{\frac{2011}{2}}$