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Question

Cho 3số thực dương a,b,c thỏa mãn $28(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})=4(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})+2013$.

Tính giá trị lớn nhất của P= $\frac{1}{\sqrt{5a^{2}+2ab+b^{2}}}+\frac{1}{\sqrt{5b^{2}+2bc+c^{2}}}+\frac{1}{\sqrt{5c^{2}+2ac+a^{2}}}$

minh nghi bai nay dai day ai lam di dai the kia biet cach ma danh thi .....

score 2 · Answer 1

$tu gia thiet va bdt A= \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} ta co$

$ 28A-2013\leq 4A<=> A\leq \frac{2013}{24}$

$ap dung bdt BCS co P\leq 3(\frac{1}{5a^{2}+2ab+b^{2}}+\frac{1}{5b^{2}+2bc+c^{2}}+\frac{1}{5c^{2}+2ac+a^{2}})$

$\frac{1}{5a^{2}+2ab+b^{2}}\leq \frac{1}{4}(\frac{1}{4a^{2}}+\frac{1}{(a+b)^{2}})\leq \frac{1}{16a^{2}}+\frac{1}{16ab}$

$ tu do tuong tu vs 2 hang tu kia dk $

$P\leq \frac{1}{16}(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})=\frac{A}{16}+\frac{28A-2013}{4.16}=\frac{A}{2}+\frac{2013}{64}\leq \frac{4697}{64}$

$vay Pmax= \frac{4697}{64} dat dk khi a=b=c=\frac{\sqrt{2013} }{6\sqrt{2} }$

score 3 · Answer 2

Từ giả thiết ta có:

$28(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})=4(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca})+2013\le4(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})+2013$

$\Rightarrow \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\le\dfrac{671}{8}$

Ta có:

$\dfrac{1}{\sqrt{5a^2+2ab+b^2}}=\dfrac{8\sqrt3}{\sqrt{671}}\sqrt{\dfrac{1}{5a^2+2ab+b^2}.\dfrac{671}{192}}$

$\le\dfrac{4\sqrt3}{\sqrt{671}}\left(\dfrac{1}{5a^2+2ab+b^2}+\dfrac{671}{192}\right)$

$\le\dfrac{4\sqrt3}{\sqrt{671}}\left[\dfrac{1}{64}\left(\dfrac{5}{a^2}+\dfrac{2}{ab}+\dfrac{1}{b^2}\right)+\dfrac{671}{192}\right]$

$=\dfrac{\sqrt3}{16\sqrt{671}}\left(\dfrac{5}{a^2}+\dfrac{2}{ab}+\dfrac{1}{b^2}\right)+\dfrac{\sqrt{671}}{16\sqrt3}$

Tương tự: $\dfrac{1}{\sqrt{5b^2+2bc+c^2}}\le\dfrac{\sqrt3}{16\sqrt{671}}\left(\dfrac{5}{b^2}+\dfrac{2}{bc}+\dfrac{1}{c^2}\right)+\dfrac{\sqrt{671}}{16\sqrt3}$

$\dfrac{1}{\sqrt{5c^2+2ca+a^2}}\le\dfrac{\sqrt3}{16\sqrt{671}}\left(\dfrac{5}{c^2}+\dfrac{2}{ca}+\dfrac{1}{a^2}\right)+\dfrac{\sqrt{671}}{16\sqrt3}$

Cộng các BĐT trên ta được:

$P\le\dfrac{6\sqrt3}{16\sqrt{671}}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+\dfrac{2\sqrt3}{16\sqrt{671}}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)+\dfrac{3\sqrt{671}}{16\sqrt3}$

$\le\dfrac{\sqrt3}{2\sqrt{671}}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+\dfrac{\sqrt{2013}}{16}\le\dfrac{\sqrt{2013}}{8}$

Vậy $\max P=\dfrac{\sqrt{2013}}{8} \Leftrightarrow a=b=c=\dfrac{2\sqrt6}{\sqrt{671}}$