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\(1=x+y+3xy\le x+y+\dfrac{3}{4}\left(x+y\right)^2\)
\(\Rightarrow3\left(x+y\right)^2+4\left(x+y\right)-4\ge0\)
\(\Rightarrow3\left(x+y+2\right)\left(x+y-\dfrac{2}{3}\right)\ge0\)
\(\Rightarrow x+y\ge\dfrac{2}{3}\) \(\Rightarrow\dfrac{1}{x+y}\le\dfrac{3}{2}\)
Đồng thời: \(x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\ge\dfrac{1}{2}.\left(\dfrac{2}{3}\right)^2=\dfrac{2}{9}\)
\(\Rightarrow-\left(x^2+y^2\right)\le-\dfrac{2}{9}\)
Từ đó ta có:
\(A=\sqrt{1-x^2}+\sqrt{1-y^2}+\dfrac{1-\left(x+y\right)}{x+y}=\sqrt{1-x^2}+\sqrt{1-y^2}+\dfrac{1}{x+y}-1\)
\(A\le\sqrt{2\left[2-\left(x^2+y^2\right)\right]}+\dfrac{1}{x+y}-1\le\sqrt{2\left(2-\dfrac{2}{9}\right)}+\dfrac{3}{2}-1=\dfrac{3+8\sqrt{2}}{6}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{3}\)
Từ giả thiết ta có:
\(x+y=3\left(\sqrt{x+1}+\sqrt{y+2}\right)\le3\sqrt{2\left(x+y+3\right)}\)
\(\Leftrightarrow P\le3\sqrt{2\left(P+3\right)}\)
\(\Leftrightarrow\left\{{}\begin{matrix}P\ge0\\18P+54\ge P^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}P\ge0\\P^2-18P-54\le0\end{matrix}\right.\)
\(\Leftrightarrow0\le P\le9+3\sqrt{15}\)
\(\Rightarrow maxP=9+3\sqrt{15}\Leftrightarrow\left(x;y\right)=\left(\dfrac{10+3\sqrt{15}}{2};\dfrac{8+3\sqrt{15}}{2}\right)\)
Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)
\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)
\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)
Tương tự ta được
\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)
\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :
\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)
\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)
\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)
\(\left(x^3+y^3\right)\left(x+y\right)=xy\left(1-x\right)\left(1-y\right)\Leftrightarrow\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)=\left(1-x\right)\left(1-y\right)\left(1\right)\)
Ta có : \(\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)\ge4xy\)
và \(\left(1-x\right)\left(1-y\right)=1-\left(x+y\right)+xy\le1-2\sqrt{xy}+xy\)
\(\Rightarrow1-2\sqrt{xy}+xy\ge4xy\Leftrightarrow0\) <\(xy\le\frac{1}{9}\)
Dễ chứng minh : \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\le\frac{1}{1+xy};\left(x,y\in\left(0;1\right)\right)\)
\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\le\sqrt{2\left(\frac{1}{1+x^2}+\frac{1}{1+y^2}\right)}\le\sqrt{2\left(\frac{2}{1+xy}\right)}=\frac{2}{\sqrt{1+xy}}\)
\(3xy-\left(x^2+y^2\right)=xy-\left(x-y\right)^2\le xy\)
\(\Rightarrow P\le\frac{2}{\sqrt{1+xy}}+xy=\frac{2}{\sqrt{1+t}}+t\), \(\left(t=xy\right)\), (0<\(t\le\frac{1}{9}\)
Xét hàm số :
\(f\left(t\right)=\frac{2}{\sqrt{t+1}}+t\) , (0<\(t\le\frac{1}{9}\)
Ta có Max \(f\left(t\right)=f\left(\frac{1}{9}\right)=\frac{6\sqrt{10}}{10}+\frac{1}{9}\), \(t\in\left(0;\frac{1}{9}\right)\)
Áp dụng BĐT Cosi cho 2018 số:
\(2017.6^{2018}.\sqrt[2017]{m}+\dfrac{\left(2a\right)^{2018}}{m}\ge2018\sqrt[2018]{\left(6^{2018}.\sqrt[2017]{m}\right)^{2017}\dfrac{\left(2a\right)^{2018}}{m}}=2018.2.6^{2017}.a\)
\(\Leftrightarrow\dfrac{\left(2a\right)^{2018}}{m}\ge2018.2.6^{2017}.a-2017.6^{2018}.\sqrt[2017]{m}\)
\(\Leftrightarrow\dfrac{2\left(2a\right)^{2018}}{m}\ge2018.4.6^{2017}.a-2017.2.6^{2018}.\sqrt[2017]{m}\)
Tương tự: \(\dfrac{2\left(2b\right)^{2018}}{n}\ge2018.4.6^{2017}.b-2017.2.6^{2018}.\sqrt[2017]{n}\)
\(\dfrac{3.c^{2018}}{p}\ge2018.3.6^{2017}.c-2017.6^{2018}.3.\sqrt[2017]{p}\)
\(\Rightarrow S\ge2018.6^{2017}\left(4a+4b+3c\right)-2017.6^{2018}\left(2\sqrt[2017]{m}+2\sqrt[2017]{n}+3\sqrt[2017]{p}\right)\)
\(\ge2018.6^{2017}.42-2017.6^{2018}.7=7.6^{2018}>6^{2018}\)
Vậy \(S>6^{2018}\)
\(3xy=x+y+1\ge3\sqrt[3]{xy}\Rightarrow xy\ge1\)
\(4xy=xy+x+y+1=x\left(y+1\right)+\left(y+1\right)=\left(x+1\right)\left(y+1\right)\)
\(P=\frac{1}{x\left(y+1\right)}+\frac{1}{y\left(x+1\right)}=\frac{2xy+x+y}{4\left(xy\right)^2}=\frac{5xy-1}{4\left(xy\right)^2}\)
Xét hiệu: \(P-1=\frac{5xy-1}{4x^2y^2}-1=\frac{\left(4xy-1\right)\left(1-xy\right)}{4x^2y^2}\le0\) với mọi \(xy\ge1\)
Vậy \(P\le1\)hay max P = 1.
Dẫu "=" xảy ra <=> x = y = 1.
Áp dụng BĐT Cauchy ta có: \(3xy\ge2\sqrt{xy}+1\Leftrightarrow xy\ge1\)
Áp dụng BĐT Cauchy ta có:
\(P=\frac{1}{x\left(y+1\right)}+\frac{1}{y\left(x+1\right)}=\frac{5xy-1}{xy\left(x+1\right)\left(y+1\right)}=\frac{5xy-1}{4\left(xy\right)^2}\), đặt t=\(\frac{1}{xy}\)
\(f\left(t\right)=\frac{5}{4}t-\frac{1}{4}t^2\)đồng biến trên (0;1] nên f(t) đạt GTLN tại t=1
Vậy GTKN của P=1 đạt được khi x=y=1
\(\sqrt{x+2017}-y^3=\sqrt{y+2017}-x^3\)
\(\Leftrightarrow\left(\sqrt{x+2017}-\sqrt{y+2017}\right)+\left(x^3-y^3\right)=0\)
\(\Leftrightarrow\dfrac{x-y}{\sqrt{x+2017}+\sqrt{y+2017}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x+2017}+\sqrt{y+2017}}+\left(x^2+xy+y^2\right)\right)=0\)
\(\Leftrightarrow x=y\)
\(\Rightarrow P=x^2-3x^2+12x-x^2+2018\)
\(=-3x^2+12x+2018=2030-3\left(x-2\right)^2\le2030\)