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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
\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{2019}\Rightarrow\dfrac{a+b}{ab}=\dfrac{1}{2019}\Rightarrow2019=\dfrac{ab}{a+b}\)
\(\dfrac{1}{a}=\dfrac{1}{2019}-\dfrac{1}{b}=\dfrac{b-2019}{2019b}\Rightarrow b-2019=\dfrac{2019b}{a}\)
\(\dfrac{1}{b}=\dfrac{1}{2019}-\dfrac{1}{a}=\dfrac{a-2019}{2019a}\Rightarrow a-2019=\dfrac{2019a}{b}\)
\(\Rightarrow\sqrt{a-2019}+\sqrt{b-2019}=\sqrt{\dfrac{2019a}{b}}+\sqrt{\dfrac{2019b}{a}}=\dfrac{\sqrt{2019}\left(a+b\right)}{\sqrt{ab}}=\sqrt{\dfrac{ab}{a+b}}.\dfrac{a+b}{\sqrt{ab}}=\sqrt{a+b}\)
\(VT=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{a^2}{b+c}\ge\frac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\frac{b^2}{\sqrt{2\left(c^2+a^2\right)}}+\frac{c^2}{\sqrt{2\left(c^2+a^2\right)}}\)
Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=\frac{y^2+z^2-x^2}{2}\\b^2=\frac{x^2+z^2-y^2}{2}\\c^2=\frac{x^2+y^2-z^2}{2}\\x+y+z=\sqrt{2019}\end{matrix}\right.\) \(\Rightarrow VT\ge\frac{1}{\sqrt{8}}\left(\frac{y^2+z^2-x^2}{x}+\frac{x^2+z^2-y^2}{y}+\frac{x^2+y^2-z^2}{z}\right)\)
\(VT\ge\frac{1}{\sqrt{8}}\left(\frac{\left(y+z\right)^2}{2x}+\frac{\left(x+z\right)^2}{2y}+\frac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\right)\)
\(VT\ge\frac{1}{\sqrt{8}}\left[\frac{\left(2x+2y+2z\right)^2}{2\left(x+y+z\right)}-\left(x+y+z\right)\right]=\frac{x+y+z}{\sqrt{8}}=\sqrt{\frac{2019}{8}}\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=\) nhiêu đó
EM tham khảo phần đầu ở link: Câu hỏi của Đinh Nguyến Nhật Minh - Toán lớp 8 - Học toán với OnlineMath
Trong 3 số a,b, c có hai số đối nhau g/s 2 số đó là a và b kho đó a=-b
=> \(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{\left(-b\right)^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=-\frac{1}{b^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{c^{2019}}\)
và \(\frac{1}{a^{2019}+b^{2019}+c^{2019}}=\frac{1}{\left(-b\right)^{2019}+b^{2019}+c^{2019}}=\frac{1}{-b^{2019}+b^{2019}+c^{2019}}=\frac{1}{c^{2019}}\)
Do đó: \(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{a^{2019}+b^{2019}+c^{2019}}\)
câu 1 tham khảo bn nhé
https://hoc24.vn/hoi-dap/question/841612.html
https://loga.vn/hoi-dap/tinh-can-1-1-2-2-1-3-2-can-1-1-2-2-1-3-2-tinh-sqrt-1-dfrac-1-2-2-dfrac-1-3-2-sqrt-1-dfrac-1-2-2-19838
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{2019}< =>\frac{2019}{a}+\frac{2019}{b}=1< =>\frac{2019}{b}=\frac{a-2019}{a}=>a-2019=\frac{2019a}{b}.\)
tương tự \(b-2019=\frac{2019b}{a}\)
=> \(\sqrt{a-2019}+\sqrt{b-2019}=\sqrt{\frac{2019a}{b}}+\sqrt{\frac{2019b}{a}}=\sqrt{2019}\left(\frac{a+b}{\sqrt{ab}}\right)\)(1)
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{2019}=>ab=2019\left(a+b\right)\)thay vào (1) ta được
\(\sqrt{2019}\left(\frac{a+b}{\sqrt{2019\left(a+b\right)}}\right)=\sqrt{a+b}\)(chứng minh xong)