5 replies to this topic

[duyanh782014]

Cho $a,b,c>0$ thỏa mãn $abc=1$.Tim Min P=$\frac{a^{3}}{a^{2}+4ab+b^{2}}+\frac{b^{3}}{b^{2}+4bc+c^{2}}+\frac{c^{3}}{c^{2}+4ac+a^{2}}$

Edited by tritanngo99, 25-09-2016 - 17:39.

[NTA1907]

Cho a,b,c>0 tm abc=1.Tim Min P=$\frac{a^{3}}{a^{2}+4ab+b^{2}}+\frac{b^{3}}{b^{2}+4bc+c^{2}}+\frac{c^{3}}{c^{2}+4ac+a^{2}}$

Áp dụng bất đẳng thức AM-GM ta có:

$a-\frac{a^{3}}{a^{2}+4ab+b^{2}}=\frac{4a^{2}b+ab^{2}}{a^{2}+ab+ab+ab+ab+b^{2}}\leq \frac{4a^{2}b+ab^{2}}{6\sqrt[6]{(ab)^{6}}}=\frac{2a}{3}+\frac{b}{6}$

$\Rightarrow \sum \frac{a^{3}}{a^{2}+4ab+b^{2}}\geq \frac{1}{6}\sum a\geq \frac{1}{6}.3\sqrt[3]{abc}=\frac{1}{2}$

Dấu "=" xảy ra$\Leftrightarrow a=b=c=1$

[hanguyen445]

Ta

 

Cho a,b,c>0 tm abc=1.Tim Min P=$\frac{a^{3}}{a^{2}+4ab+b^{2}}+\frac{b^{3}}{b^{2}+4bc+c^{2}}+\frac{c^{3}}{c^{2}+4ac+a^{2}}$

 Ta có:

 \[\left( {x + y + z} \right)\left( {\frac{{{a^3}}}{x} + \frac{{{b^3}}}{y} + \frac{{{c^3}}}{z}} \right)\left( {1 + 1 + 1} \right) \ge {\left( {a + b + c} \right)^3}\]

\[ \Leftrightarrow \frac{{{a^3}}}{x} + \frac{{{b^3}}}{y} + \frac{{{c^3}}}{z} \ge \frac{{{{\left( {a + b + c} \right)}^3}}}{{3\left( {a + y + z} \right)}}\]

Áp dụng BĐT trên. Ta có:

\[P = \frac{{{a^3}}}{{{a^2} + 4ab + {b^2}}} + \frac{{{b^3}}}{{{b^2} + 4bc + {c^2}}} + \frac{{{c^3}}}{{{c^2} + 4ac + {a^2}}}\]

\[ \ge \frac{{{{\left( {a + b + c} \right)}^3}}}{{3\sum {\left( {{a^2} + {b^2} + 4ab} \right)} }}\]

\[ = \frac{{{{\left( {a + b + c} \right)}^3}}}{{6{{\left( {a + b + c} \right)}^2}}} = \frac{{a + b + c}}{6} \ge \frac{{3\sqrt[3]{{abc}}}}{6} = \frac{1}{2}\]

[basketball123]

Cho a,b,c>0 tm abc=1.Tim Min P=$\frac{a^{3}}{a^{2}+4ab+b^{2}}+\frac{b^{3}}{b^{2}+4bc+c^{2}}+\frac{c^{3}}{c^{2}+4ac+a^{2}}$

$\frac{a^{3}}{a^{2}+4ab+b^{2}}+\frac{b^{3}}{b^{2}+4bc+c^{2}}+\frac{c^{3}}{c^{2}+4ca+a^{2}}\geq \frac{1}{3}(\frac{a^{3}}{a^{2}+b^{2}}+\frac{b^{3}}{b^{2}+c^{2}}+\frac{c^{3}}{c^{2}+a^{2}})\geq \frac{1}{6}(a+b+c)\geq \frac{1}{2}$

Đẳng thức xảy ra $\Leftrightarrow a=b=c=1$

[sharker]

$\frac{a^3}{a^2+4ab+b^2}\geq -\frac{1}{6}.b+\frac{1}{3}.a

\Leftrightarrow \frac{(a-b)^2(4a+b)}{6(a^2+4ab+b^2)}

\rightarrow \sum \frac{a^3}{a^2+4ab+b^2}\geq \frac{1}{6}(a+b+c)\geq \frac{1}{6}.3\sqrt[3]{abc}=\frac{1}{2}$

Edited by sharker, 25-09-2016 - 20:06.

[sharker]

$\frac{a^3}{a^2+4ab+b^2}\geq -\frac{1}{6}.b+\frac{1}{3}.a \Leftrightarrow \frac{(a-b)^2(4a+b)}{6(a^2+4ab+b^2)} \rightarrow \sum \frac{a^3}{a^2+4ab+b^2}\geq \frac{1}{6}(a+b+c)\geq \frac{1}{6}.3\sqrt[3]{abc}=\frac{1}{2}$