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Ta có: \(x+y\ge2\sqrt{xy}\Rightarrow3xy\ge2\sqrt{xy}+1\Rightarrow3xy-2\sqrt{xy}-1\ge0\)
\(\Rightarrow\left(3\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)\ge0\Rightarrow\sqrt{xy}-1\ge0\) (do \(3\sqrt{xy}+1>0\) )
\(\Rightarrow\sqrt{xy}\ge1\Rightarrow xy\ge1\Rightarrow1-xy\le0\)
\(P=\dfrac{y\left(x+1\right)+x\left(y+1\right)}{xy\left(x+1\right)\left(y+1\right)}=\dfrac{2xy+x+y}{xy\left(xy+x+y+1\right)}\)
\(\Rightarrow P=\dfrac{2xy+3xy-1}{xy\left(xy+3xy\right)}=\dfrac{5xy-1}{4\left(xy\right)^2}=\dfrac{-4\left(xy\right)^2+5xy-1}{4\left(xy\right)^2}+1\)
\(\Rightarrow P=\dfrac{\left(1-xy\right)\left(4xy+1\right)}{4\left(xy\right)^2}+1\)
Do \(\left\{{}\begin{matrix}1-xy\le0\\4xy+1>0\\4\left(xy\right)^2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\left(1-xy\right)\left(4xy+1\right)}{4\left(xy\right)^2}\le0\)
\(\Rightarrow P\le0+1=1\Rightarrow P_{max}=1\) khi \(x=y=1\)
\(1=x+y+3xy\le x+y+\dfrac{3}{4}\left(x+y\right)^2\)
\(\Rightarrow3\left(x+y\right)^2+4\left(x+y\right)-4\ge0\)
\(\Rightarrow3\left(x+y+2\right)\left(x+y-\dfrac{2}{3}\right)\ge0\)
\(\Rightarrow x+y\ge\dfrac{2}{3}\) \(\Rightarrow\dfrac{1}{x+y}\le\dfrac{3}{2}\)
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 \(x=y=\dfrac{1}{3}\)
Lời giải:
Sử dụng bổ đề: Với \(a,b>0\Rightarrow a^3+b^3\geq ab(a+b)\)
BĐT đúng vì nó tương đương với \((a-b)^2(a+b)\geq 0\) (luôn đúng)
Áp dụng vào bài toán:
\(P\leq \frac{1}{x^3yz(y+z)+1}+\frac{1}{y^3xz(x+z)+1}+\frac{1}{z^3xy(x+y)+1}\)
\(\Leftrightarrow P\leq \frac{1}{x^2(y+z)+xyz}+\frac{1}{y^2(x+z)+xyz}+\frac{1}{z^2(x+y)+xyz}\)
\(\Leftrightarrow P\leq \frac{1}{x(xy+yz+xz)}+\frac{1}{y(xy+yz+xz)}+\frac{1}{z(xy+yz+xz)}=\frac{xy+yz+xz}{xy+yz+xz}=1\)
Vậy \(P_{\max}=1\Leftrightarrow x=y=z=1\)
\(\left(x^3+y^3\right)\left(x+y\right)=xy\left(1-x\right)\left(1-y\right)\Leftrightarrow\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)=\left(1-x\right)\left(1-y\right)\left(1\right)\)
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<\(t\le\frac{1}{9}\)
Từ giả thiết ta có:
\(x+y=3\left(\sqrt{x+1}+\sqrt{y+2}\right)\le3\sqrt{2\left(x+y+3\right)}\)
\(\Leftrightarrow P\le3\sqrt{2\left(P+3\right)}\)
\(\Leftrightarrow\left\{{}\begin{matrix}P\ge0\\18P+54\ge P^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}P\ge0\\P^2-18P-54\le0\end{matrix}\right.\)
\(\Leftrightarrow0\le P\le9+3\sqrt{15}\)
\(\Rightarrow maxP=9+3\sqrt{15}\Leftrightarrow\left(x;y\right)=\left(\dfrac{10+3\sqrt{15}}{2};\dfrac{8+3\sqrt{15}}{2}\right)\)
đặt \(\left(a;b;c\right)=\left(\sqrt{\frac{yz}{x}};\sqrt{\frac{zx}{y}};\sqrt{\frac{xy}{z}}\right)\)\(\Rightarrow\)\(a^2+b^2+c^2=1\)
\(A=\Sigma\frac{1}{1-ab}=\Sigma\frac{2ab}{2\left(a^2+b^2+c^2\right)-2ab}+3\le\frac{1}{2}\Sigma\frac{\left(a+b\right)^2}{b^2+c^2+c^2+a^2}\)
\(\le\frac{1}{2}\Sigma\left(\frac{a^2}{c^2+a^2}+\frac{b^2}{b^2+c^2}\right)=\frac{9}{2}\)
dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{3}\)
\(Ta \) \(có : x^2 +y^2 +xy = 1\)
\(\Leftrightarrow\)\(xy = 1 - x^2 - y^2\)
\(Thay \) \(xy = 1 - x^2 - y^2 \) \(vào \) \(P , ta \) \(được :\)
\(P = 1 - x^2 -y^2\)
\(P = 1 - ( x^2 +y^2 )\)
\(P = - ( x^2 +y^2 )+ 1\)\(\le\)\(1\)
\(Dấu "=" xảy \) \(ra\) \(\Leftrightarrow\)\(x^2+y^2 =0\)
\(\Leftrightarrow\)\(x = 0 \) \(và\) \(y = 0\)
\(Max \) \(P = 1 \)\(\Leftrightarrow\)\(x = 0 ; y = 0\)
\(3xy=x+y+1\ge3\sqrt[3]{xy}\Rightarrow xy\ge1\)
\(4xy=xy+x+y+1=x\left(y+1\right)+\left(y+1\right)=\left(x+1\right)\left(y+1\right)\)
\(P=\frac{1}{x\left(y+1\right)}+\frac{1}{y\left(x+1\right)}=\frac{2xy+x+y}{4\left(xy\right)^2}=\frac{5xy-1}{4\left(xy\right)^2}\)
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