Cho \(0\le x,y\le\dfrac{1}{2}\).CM: \(\dfrac{\sqrt{x}}{1+y}+\dfrac{\sqrt{y}}{1+x}\le\dfrac{2\sqrt{2}}{3}\)
Cho x,y thỏa mãn x,y thuộc R và 0\(\le x,y\le\dfrac{1}{2}\) chứng minh rằng \(\dfrac{\sqrt{x}}{1+y}+\dfrac{\sqrt{y}}{1+x}\le\dfrac{2\sqrt{2}}{3}\)
C.hóa \(x+y=1\) và dùng C-S:
\(VT^2\le\frac{2x}{\left(y+1\right)^2}+\frac{2y}{\left(x+1\right)^2}\le\frac{8}{9}=VP^2\)
\(BDT\Leftrightarrow\frac{x}{\left(2-x\right)^2}+\frac{y}{\left(2-y\right)^2}\le\frac{4}{9}\left(1\right)\)
Ta có BĐT phụ \(\frac{x}{\left(2-x\right)^2}\le\frac{20}{27}x-\frac{4}{27}\)
\(\Leftrightarrow-\frac{\left(2x-1\right)^2\left(5x-16\right)}{27\left(x-2\right)^2}\le0\) *Đúng*
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT_{\left(1\right)}\le\frac{20}{27}\left(x+y\right)-\frac{4}{27}\cdot2=\frac{4}{9}=VP_{\left(1\right)}\)
"=" khi \(x=y=\frac{1}{2}\)
Cho x,y thỏa mãn x,y thuộc R và 0\(\le x,y\le\dfrac{1}{2}\) chứng minh rằng \(\dfrac{\sqrt{x}}{1+y}+\dfrac{\sqrt{y}}{1+x}\le\dfrac{2\sqrt{2}}{3}\)
༺ ๖ۣۜPhạm ✌Tuấn ✌Kiệτ ༻Tâm đường tròn ở đâu
Cho x,y thỏa mãn x,y thuộc R và 0\(\le x,y\le\dfrac{1}{2}\) chứng minh rằng \(\dfrac{\sqrt{x}}{1+y}+\dfrac{\sqrt{y}}{1+x}\le\dfrac{2\sqrt{2}}{3}\)
Bạn tham khảo lời giải tại đây:
https://hoc24.vn/cau-hoi/voi-0-xy-dfrac12-chung-minhdfracsqrtxy1dfracsqrtyx1-dfrac2sqrt23.461470553384
Cho x,y,z > 0 và \(x+y+z\le\dfrac{3}{2}\). CMR :
\(\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\ge\dfrac{3}{2}\sqrt{17}\)
Cho x,y,z>0 và x+y+z = xyz
CMR
\(\dfrac{1}{\sqrt{x^2+1}}+\dfrac{1}{\sqrt{y^2+1}}+\dfrac{1}{\sqrt{z^2+1}}\le\dfrac{3}{2}\)
\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow ab+bc+ca=1\)
Đặt vế trái là P, ta có:
\(P=\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\)
\(P=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(P=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)+\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)+\dfrac{1}{2}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\) hay \(x=y=z=\sqrt{3}\)
cho x,y,z>0 thỏa mãn \(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\).CMR \(\sqrt{x}+\sqrt{y}+\sqrt{z}\le\dfrac{3}{2}\sqrt{xyz}\)
Giả thiết thiếu rồi em, chỗ \(\dfrac{1}{x+1}+...\) thiếu đoạn sau nữa
Cho x,y,z>0
\(CM:\sqrt{\dfrac{x}{z+3x}}+\sqrt{\dfrac{y}{x+3y}}+\sqrt{\dfrac{z}{y+3z}}\le\dfrac{3}{2}\)
Đặt vế trái là P, ta có:
\(P\le\sqrt{3\left(\dfrac{x}{z+3x}+\dfrac{y}{x+3y}+\dfrac{z}{y+3z}\right)}\)
Nên ta chỉ cần chứng mình: \(\sqrt{3\left(\dfrac{x}{z+3x}+\dfrac{y}{x+3y}+\dfrac{z}{y+3z}\right)}\le\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x}{z+3x}+\dfrac{y}{x+3y}+\dfrac{z}{y+3z}\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3x}{z+3x}-1+\dfrac{3y}{x+3y}-\dfrac{3z}{y+3z}-1\le\dfrac{9}{4}-3\)
\(\Leftrightarrow\dfrac{z}{z+3x}+\dfrac{x}{x+3y}+\dfrac{y}{y+3z}\ge\dfrac{3}{4}\)
BĐT trên đúng do:
\(\dfrac{z}{z+3x}+\dfrac{x}{x+3y}+\dfrac{y}{y+3z}=\dfrac{z^2}{z^2+3zx}+\dfrac{x^2}{x^2+3xy}+\dfrac{y^2}{y^2+3yz}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+xy+yz+zx}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\dfrac{1}{3}\left(x+y+z\right)^2}=\dfrac{3}{4}\)
cho x,y,z>0 thỏa mãn \(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}=1\\\).CMR
\(\sqrt{x}+\sqrt{y}+\sqrt{z}\le\dfrac{3}{2}\sqrt{xyz}\)
Đặt \(\left(\dfrac{1}{\sqrt{x}};\dfrac{1}{\sqrt{y}};\dfrac{1}{\sqrt{z}}\right)=\left(a;b;c\right)\Rightarrow\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}=1\)
Ta cần chứng minh: \(ab+bc+ca\le\dfrac{3}{2}\)
Thật vậy, ta có:
\(1=\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3}\)
\(\Rightarrow a^2+b^2+c^2+3\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Rightarrow ab+bc+ca\le\dfrac{3}{2}\) (đpcm)
Cho x,y,z>0;\(x+y+z\le\dfrac{3}{2}\).CMR
\(\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\ge\dfrac{3}{2}\sqrt{17}\)
Mn giúp e với (có thể dùng bunhiacopxki nhé mn)
Xài Bunhiacopxki thì bài này sẽ hơi dài:
Đặt vế trái là P
Ta có:
\(\left(\dfrac{1}{4}+4\right)\left(x^2+\dfrac{1}{x^2}\right)\ge\left(\dfrac{x}{2}+\dfrac{2}{x}\right)^2\)
\(\Leftrightarrow\dfrac{17}{4}\left(x^2+\dfrac{1}{x^2}\right)\ge\left(\dfrac{x}{2}+\dfrac{2}{x}\right)^2\)
\(\Rightarrow\sqrt{x^2+\dfrac{1}{x^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{x}{2}+\dfrac{2}{x}\right)\)
Tương tự:
\(\sqrt{y^2+\dfrac{1}{y^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{y}{2}+\dfrac{2}{y}\right)\) ; \(\sqrt{z^2+\dfrac{1}{z^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{z}{2}+\dfrac{2}{z}\right)\)
Cộng vế: \(P\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{x}{2}+\dfrac{y}{2}+\dfrac{z}{2}+\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}\right)\)
\(P\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right)\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+\dfrac{36}{x+y+z}\right)\)
\(P\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+\dfrac{9}{4\left(x+y+z\right)}+\dfrac{135}{4\left(x+y+z\right)}\right)\)
\(P\ge\dfrac{1}{\sqrt{17}}\left(2\sqrt{\dfrac{9\left(x+y+z\right)}{4\left(x+y+z\right)}}+\dfrac{135}{4.\dfrac{3}{2}}\right)=\dfrac{3}{2}\sqrt{17}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)