6 replies to this topic

[duyanh782014]

Cho a,b,c>0 thỏa mãn a+b+c=1.Tìm min P=$\frac{a}{(b+c)^{2}}+\frac{b}{(c+a)^{2}}+\frac{c}{(a+b)^{2}}$

[duyanh782014]

Thôi mình giải đc r k cần làm đâu

[hanguyen445]

\[{\left( {a + b + c} \right)^2} = 1 \geqslant 3\left( {ab + bc + ac} \right)\]

\[ + \frac{a}{{{{\left( {b + c} \right)}^2}}} + \frac{{27a\left( {b + c} \right)}}{8} + \frac{{27a\left( {b + c} \right)}}{8}\]

\[ \geqslant 3\sqrt[3]{{\frac{{{a^3}{{.27}^2}}}{{{8^2}}}}} = \frac{{27a}}{4}\]

\[P = \sum {\frac{a}{{{{\left( {b + c} \right)}^2}}} \geqslant \frac{{27}}{4}\left( {a + b + c} \right) - \frac{{27}}{2}\left( {ab + bc + ac} \right)} \]

\[ \geqslant \frac{{27}}{4} - \frac{9}{2}{\left( {a + b + c} \right)^2} = \frac{{27}}{4} - \frac{9}{2} = \frac{9}{4}\]

 

Cho a,b,c>0 thỏa mãn a+b+c=1.Tìm min P=$\frac{a}{(b+c)^{2}}+\frac{b}{(c+a)^{2}}+\frac{c}{(a+b)^{2}}$

[Element hero Neos]

\[{\left( {a + b + c} \right)^2} = 1 \geqslant 3\left( {ab + bc + ac} \right)\]

\[ + \frac{a}{{{{\left( {b + c} \right)}^2}}} + \frac{{27a\left( {b + c} \right)}}{8} + \frac{{27a\left( {b + c} \right)}}{8}\]

\[ \geqslant 3\sqrt[3]{{\frac{{{a^3}{{.27}^2}}}{{{8^2}}}}} = \frac{{27a}}{4}\]

\[P = \sum {\frac{a}{{{{\left( {b + c} \right)}^2}}} \geqslant \frac{{27}}{4}\left( {a + b + c} \right) - \frac{{27}}{2}\left( {ab + bc + ac} \right)} \]

\[ \geqslant \frac{{27}}{4} - \frac{9}{2}{\left( {a + b + c} \right)^2} = \frac{{27}}{4} - \frac{9}{2} = \frac{9}{4}\]

Cách khác

Ta có

$\frac{a}{(1-a)^2}-\frac{9}{2}a+\frac{3}{4}=\frac{(3a-1)^2(3-2a)}{4(1-a)^2}\geq 0$

TT, cộng lại có

$P\geq\frac{9}{2}(a+b+c)-\frac{3}{4}.3=\frac{9}{4}$

Vậy ...

[kienvuhoang]

Áp dụng BCS ta có

$[\frac{a}{(b+c)^{2}}+\frac{b}{(c+a)^{2}}+\frac{c}{(a+b)^{2}}](a+b+c)

\geq (\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})^{2}

\geq \frac{9}{4}\Rightarrow P\geq \frac{9}{4}$

Đẳng thức xảy ra kvck a=b=c=1/3

[doanminhhien127]

bài này có làm theo cách bunhiacopxki đc ko. Nếu có thì các bạn chỉ giúp mình vs. Cảm ơn nhiều

[KietLW9]

Cho a,b,c>0 thỏa mãn a+b+c=1.Tìm min P=$\frac{a}{(b+c)^{2}}+\frac{b}{(c+a)^{2}}+\frac{c}{(a+b)^{2}}$

Áp dụng Cô-si, ta có: $$a(b+c)^2+b(c+a)^2+c(a+b)^2=\frac{1}{2}.2a(b+c)(b+c)+\frac{1}{2}.2b(c+a)(c+a)+\frac{1}{2}.2c(a+b)(a+b)\leqslant\frac{1}{2}.\frac{8(a+b+c)^3}{27}+ \frac{1}{2}.\frac{8(a+b+c)^3}{27}+\frac{1}{2}.\frac{8(a+b+c)^3}{27}=\frac{4}{9}(a+b+c)^3$$

$\Rightarrow \sum_{cyc}\frac{a}{(b+c)^2}= \sum_{cyc}\frac{a^2}{a(b+c)^2}\geqslant \frac{(a+b+c)^2}{\sum_{cyc}a(b+c)^2} \geqslant \frac{(a+b+c)^2}{\frac{4}{9}(a+b+c)^3} =\frac{9}{4(a+b+c)}=\frac{9}{4}$

Đẳng thức xảy ra khi $a = b = c =\frac{1}{3} $