Vô lí vì a+b+c=0\(\Rightarrow\frac{5}{a+b+c}\)không có đáp án

\(VT\leΣ\frac{1}{a^2+b^2+1}\le\frac{a^2+b^2+c^2+6}{\left(a+b+c\right)^2}\le\frac{\left(Σa\right)^2}{\left(Σa\right)^2}=1=VP\)

Bạn giải rõ ra được không

bạn thế 2019=a+b+c de thoi ma

Ta có: \(2019a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(c+a\right)\ge\left(\sqrt{ab}+\sqrt{ac}\right)^2\)

\(\Rightarrow a+\sqrt{2019a+bc}\ge a+\sqrt{ab}+\sqrt{bc}=\sqrt{a}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

\(\Rightarrow\frac{a}{a+\sqrt{2019a+bc}}\le\frac{a}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Tương tự cộng vào suy ra điều phải chứng minh

\(-1=-\left(a^2+b^2+c^2\right)=>-1\le2\left(ab+bc+ca\right).\\ < =>\left(a+b+c\right)^2\ge0.\)
Luôn đúng .
\(a^2+b^2+c^2=1\ge ab+bc+ca\)

\(DPCM\Leftrightarrow P=a^2\left(b-c\right)+b^2\left(c-b\right)+c^2\left(1-c\right)\le\frac{108}{529}\)

Ta có: \(0\le a\le b\le c\le1\Rightarrow a^2\left(b-c\right)\le0\left(1\right)\)

\(b^2\left(c-b\right)=4.\frac{b}{2}.\frac{b}{2}.\left(c-b\right)\le4\left(\frac{\frac{b}{2}+\frac{b}{2}+c-b}{3}\right)^3=\frac{4c^3}{27}\)

\(\Rightarrow P\le\frac{4c^3}{27}+c^2\left(1-c\right)=c^2\left(1-\frac{23c}{27}\right)=\frac{23c}{54}.\frac{23c}{54}\left(1-\frac{23c}{27}\right).\frac{54^2}{23^2}\)

Tiếp

\(\le\left(\frac{\frac{23c}{54}+\frac{23c}{54}+1-\frac{23c}{27}}{3}\right)^3.\frac{54^2}{23^2}=\frac{1}{27}.\frac{54^2}{23^2}=\frac{108}{529}\)

Dấu bằng xảy ra\(\Leftrightarrow\hept{\begin{cases}a^2\left(b-c\right)=0\\\frac{b}{2}=c-b\\\frac{23c}{54}=1-\frac{23c}{27}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=0\\b=\frac{2}{3}c\\c=\frac{18}{23}\end{cases}}\)