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There have been 2 items by holobleep (Search limited from 22-05-2020)
Posted by holobleep on 30-03-2021 - 12:30 in Phương trình, hệ phương trình và bất phương trình
Giải hệ sau
\[\left\{ \begin{array}{l} \left( {x + y} \right)\left( {x + 1} \right)\left( {y + 1} \right) = 8\\ 7{y^3} + 6xy\left( {x + 2y} \right) = 25 \end{array} \right.\]
Posted by holobleep on 30-03-2021 - 12:28 in Bất đẳng thức và cực trị
Có:$\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c}=\frac{1}{2}(\frac{a}{2-a}+1+\frac{b}{2-b}+1+\frac{c}{2-c}+1)=\frac{1}{2}(\frac{a^{2}}{2a-a^{2}}+\frac{b^{2}}{2b-b^{2}}+\frac{c^{2}}{2c-c^{2}})+\frac{3}{2}\geqslant \frac{1}{2}.\frac{(a+b+c)^{2}}{2(a+b+c)-a^{2}-b^{2}-c^{2}}+\frac{3}{2}$ Ta sẽ chứng minh:$\frac{1}{2}.\frac{(a+b+c)^{2}}{2(a+b+c)-a^{2}-b^{2}-c^{2}}+\frac{3}{2}\geqslant 3\Leftrightarrow (a+b+c)^{2}+9 \geqslant 6(a+b+c)\Leftrightarrow (a+b+c-3)^{2}\geqslant 0$( luôn đúng) suy ra đpcm Dấu bằng xảy ra khi a=b=c=1
Có:$\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c}=\frac{1}{2}(\frac{a}{2-a}+1+\frac{b}{2-b}+1+\frac{c}{2-c}+1)=\frac{1}{2}(\frac{a^{2}}{2a-a^{2}}+\frac{b^{2}}{2b-b^{2}}+\frac{c^{2}}{2c-c^{2}})+\frac{3}{2}\geqslant \frac{1}{2}.\frac{(a+b+c)^{2}}{2(a+b+c)-a^{2}-b^{2}-c^{2}}+\frac{3}{2}$
Ta sẽ chứng minh:$\frac{1}{2}.\frac{(a+b+c)^{2}}{2(a+b+c)-a^{2}-b^{2}-c^{2}}+\frac{3}{2}\geqslant 3\Leftrightarrow (a+b+c)^{2}+9 \geqslant 6(a+b+c)\Leftrightarrow (a+b+c-3)^{2}\geqslant 0$( luôn đúng) suy ra đpcm
Dấu bằng xảy ra khi a=b=c=1