cho ab+bc+ac=3abc. chứng minh a/a^2+bc + b/ b^2+ac + c/ c^2 +ab <= 3/2 27/09/2021 Bởi Brielle cho ab+bc+ac=3abc. chứng minh a/a^2+bc + b/ b^2+ac + c/ c^2 +ab <= 3/2
Đặt `Q=a/(a^2+bc)+b/(b^2+ac)+c/(c^2+ab)` Áp dụng BĐT Co-si ta có: `+)a^2+bc>=2a` $\sqrt[]{ab}$ `=>a/(a^2+bc) `$\leq$ $\frac{1}{2\sqrt[]{bc}}$ ` <= 1/4 ( 1/b +1/c)` `+)b^2+ca>=2b` $\sqrt[]{ca}$ `=>b/(b^2+ca) `$\leq$ $\frac{1}{2\sqrt[]{ca}}$ ` <= 1/4 ( 1/a +1/c)` `+)c^2+ab>=2c` $\sqrt[]{ab}$ `=>c/(c^2+ab) ` $\leq$ $\frac{1}{2\sqrt[]{ab}}$ ` <= 1/4 ( 1/a +1/b)` `=>Q=< 1/2(1/a+1/b+1/c)=1/2 * (ab+bc+ca)/(abc)=3/2` `=>a/(a^2+bc )+ b/ (b^2+ac )+ c/ (c^2 +ab) <= 3/2(đpcm)` Bình luận
Đặt `Q=a/(a^2+bc)+b/(b^2+ac)+c/(c^2+ab)`
Áp dụng BĐT Co-si ta có:
`+)a^2+bc>=2a` $\sqrt[]{ab}$
`=>a/(a^2+bc) `$\leq$ $\frac{1}{2\sqrt[]{bc}}$ ` <= 1/4 ( 1/b +1/c)`
`+)b^2+ca>=2b` $\sqrt[]{ca}$
`=>b/(b^2+ca) `$\leq$ $\frac{1}{2\sqrt[]{ca}}$ ` <= 1/4 ( 1/a +1/c)`
`+)c^2+ab>=2c` $\sqrt[]{ab}$
`=>c/(c^2+ab) ` $\leq$ $\frac{1}{2\sqrt[]{ab}}$ ` <= 1/4 ( 1/a +1/b)`
`=>Q=< 1/2(1/a+1/b+1/c)=1/2 * (ab+bc+ca)/(abc)=3/2`
`=>a/(a^2+bc )+ b/ (b^2+ac )+ c/ (c^2 +ab) <= 3/2(đpcm)`