Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Sigma\frac{x^3}{y^2}=\Sigma\frac{x}{y^2}\left(x-y\right)^2+\frac{\Sigma z\left(x^3-yz^2\right)^2}{xyz\left(x+y+z\right)}+\Sigma\frac{x^2}{y}\ge\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\)
vì x+y+z=1nên
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\)\(\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}\)\(=3+\left(\frac{x}{y}+\frac{y}{z}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\)=\(3+\frac{x^2+y^2}{xy}+\frac{y^2+z^2}{yz}+\frac{x^2+z^2}{xz}\)
nen \(\frac{xy}{x^2+y^2}+\frac{yz}{y^2+z^2}+\frac{xz}{x^2+z^2}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) =\(\left(\frac{xy}{x^2+y^2}+\frac{x^2+y^2}{4xy}\right)+\left(\frac{yz}{y^2+z^2}+\frac{y^2+z^2}{4yz}\right)+\left(\frac{xz}{x^2+z^2}+\frac{x^2+z^2}{xz}\right)+\frac{3}{4}\)
\(\ge2.\frac{1}{2}+\frac{2.1}{2}+\frac{2.1}{2}+\frac{3}{4}=\frac{15}{4}\)(dpcm)
dau = xay ra khi x=y=z=1/3
Đặt \(H=\frac{xz}{y^2+yz}+\frac{y^2}{zx+yz}+\frac{x+2z}{x+z}\)
\(=\frac{1}{\frac{y^2}{xz}+\frac{yz}{xz}}+\frac{1}{\frac{zx}{y^2}+\frac{yz}{y^2}}+\frac{x+z+z}{x+z}\)
\(=\frac{1}{\frac{y^2}{zx}+\frac{y}{x}}+\frac{1}{\frac{zx}{y^2}+\frac{z}{y}}+\frac{1}{\frac{x}{z}+1}+1\)
Đặt \(\frac{x}{y}=a;\frac{y}{z}=b\Rightarrow ab=\frac{x}{z}\ge1\)
Khi đó \(H=\frac{1}{\frac{b}{a}+\frac{1}{a}}+\frac{1}{\frac{a}{b}+\frac{1}{b}}+\frac{1}{ab+1}+1\)
\(=\frac{a}{b+1}+\frac{b}{a+b}+\frac{1}{ab+1}+1\)
Ta cần chứng minh \(U=\frac{a}{b+c}+\frac{b}{a+b}+\frac{1}{ab+1}\ge\frac{3}{2}\)
\(\Leftrightarrow\left(\frac{a}{b+1}+1\right)+\left(\frac{b}{a+1}+1\right)+\frac{1}{ab+1}\ge\frac{7}{2}\)
\(\Leftrightarrow\frac{a+b+1}{b+1}+\frac{a+b+1}{a+1}+\frac{1}{ab+1}\ge\frac{7}{2}\)
\(\Leftrightarrow\left(a+b+1\right)\left(\frac{1}{b+1}+\frac{1}{a+1}\right)+\frac{1}{ab+1}\ge\frac{7}{2}\)
Khi đó \(Y=\left(a+b+1\right)\left(\frac{1}{a+1}+\frac{1}{b+1}\right)+\frac{1}{ab+1}\)
\(\ge\left(a+b+1\right)\cdot\frac{4}{a+b+2}+\frac{1}{ab+1}\)
\(\ge\frac{4\left(a+b+1\right)}{a+b+2}+\frac{1}{\frac{\left(a+b\right)^2}{4}+1}\)
Đặt \(t=a+b\ge2\sqrt{ab}\ge2\)
Ta cần chứng minh \(\frac{4\left(t+1\right)}{t+2}+\frac{1}{\frac{t^2}{4}+1}\ge\frac{7}{2}\)
\(\Leftrightarrow\frac{\left(t-2\right)^3}{2\left(t+2\right)\left(t^2+4\right)}\ge0\) ( đúng )
Vậy ta có đpcm.
ta có:
\(\frac{xz}{y^2+yz}+\frac{y^2}{xz+yz}+\frac{z+2z}{z+x}=\frac{\frac{xz}{yz}}{\frac{y^2}{yz}+1}+\frac{\frac{y^2}{yz}}{\frac{xz}{yz}+1}+\frac{1+\frac{2z}{x}}{1+\frac{z}{x}}\)\(=\frac{\frac{x}{y}}{\frac{y}{z}+1}+\frac{\frac{y}{z}}{\frac{x}{y}+1}+\frac{1+\frac{2z}{x}}{1+\frac{z}{x}}=\frac{a^2}{b^2+1}+\frac{b^2}{a^2+1}+\frac{1+2c^2}{1+c^2}\)
trong đó \(a^2=\frac{x}{y};b^2=\frac{y}{z};c^2=\frac{z}{x}\left(a;b;c>0\right)\)
Nhận xét rằng \(a^2\cdot b^2=\frac{x}{z}=\frac{1}{c^2}\ge1\)(do x>=z)
Xét \(\frac{a^2}{b^2+1}+\frac{b^2}{a^2+1}+\frac{c^2}{ab+1}\)\(=\frac{a^2\left(a^2+1\right)\left(ab+1\right)+b^2\left(b^2+1\right)\left(ab+1\right)-2aba^2\left(a^2+1\right)\left(b^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\)
\(=\frac{ab\left(a^2-b^2\right)+\left(a-b\right)\left(a^3-b^3\right)+\left(a-b\right)^2}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)
Do đó: \(\frac{a^2}{b^2+1}+\frac{b^2}{a^2+1}\ge\frac{2ab}{ab+1}=\frac{\frac{2}{c}}{\frac{1}{c}+1}=\frac{2}{1+c}\left(1\right)\)đẳng thức xảy ra <=> a=b
khi đó:
\(\frac{2}{1+c}+\frac{1+2c^2}{c^2+1}-\frac{5}{2}=\frac{2\left[2\left(1+c^2\right)+\left(1+c\right)\left(1+2c^2\right)\right]-5\left(1+c\right)\left(1+c^2\right)}{2\left(1+c\right)\left(1+c^2\right)}\)
\(=\frac{1-3c+3c^2-c^3}{2\left(1+c\right)\left(1+c^2\right)}=\frac{\left(1-c\right)^3}{2\left(1+c\right)\left(1+c^2\right)}\ge0\)(do c=<1) (2)
Từ (1) và (2) => đpcm
Đẳng thức xảy ra <=> a=b, c=1 <=> x=y=z
\(\left(1.x+9.\frac{1}{y}\right)^2\le\left(1^2+9^2\right)\left(x^2+\frac{1}{y^2}\right)\Rightarrow\sqrt{x^2+\frac{1}{y^2}}\)
\(\ge\frac{1}{\sqrt{82}}\left(x+\frac{9}{y}\right)\)
\(TT:\sqrt{y^2+\frac{1}{z^2}}\ge\frac{1}{\sqrt{82}}\left(x+\frac{9}{z}\right);\sqrt{z^2+\frac{1}{x^2}}\ge\frac{1}{\sqrt{82}}\left(z+\frac{9}{x}\right)\)
\(S\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\right)\)
\(\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{81}{x+y+z}\right)\)
\(=\frac{1}{\sqrt{82}}\left[\left(x+y+z+\frac{1}{x+y+z}\right)+\frac{80}{x+y+z}\right]\ge\sqrt{82}\)
Ta có \(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge\frac{x+y+z}{2}\)
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z\ge\frac{x+y+z}{2}+x+y+z\)
\(\Rightarrow x\left(\frac{x}{y+z}+1\right)+y\left(\frac{y}{x+z}+1\right)+z\left(\frac{z}{x+y}+1\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Rightarrow x\left(\frac{x+y+z}{y+z}\right)+y\left(\frac{y+x+z}{x+z}\right)+z\left(\frac{z+x+y}{x+y}\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Rightarrow\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge\frac{3}{2}\) (Theo BĐT Nesbitt )
\(\Rightarrow\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge\frac{3}{2}\) (đpcm)
\(BDT\Leftrightarrow\text{∑}\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)\ge\frac{21}{2}\)
Mà \(\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge2\)(dùng AM-GM giải quyết chỗ này)
Vậy ta cần chứng minh \(\frac{y^2}{z^2}+\frac{z^2}{y^2}+\frac{z^2}{x^2}+\frac{x^2}{z^2}\ge\frac{17}{2}\)
\(\Leftrightarrow\frac{y^2}{z^2}+\frac{x^2}{z^2}\ge\frac{1}{2}\left(\frac{x}{z}+\frac{y}{z}\right)^2\)
\(\Leftrightarrow\frac{z^2}{y^2}+\frac{z^2}{x^2}\ge\frac{1}{2}\left(\frac{4z}{x+y}\right)^2\)
Đặt \(a=\frac{z}{x+y}\ge1\),ta chứng minh \(\frac{1}{2a^2}+8a^2\ge\frac{17}{2}\)
Dễ thấy BĐT này đúng.Vậy ta có đpcm
Giờ mình lội lại noti mới nhìn thấy bài tag. Không biết bạn còn cần không nhưng mình vẫn post lời giải để mọi người tham khảo:
Ta có:
\(\frac{xz}{y^2+yz}+\frac{y^2}{xz+yz}+\frac{x+2z}{x+z}=\frac{xz}{y^2+yz}+\frac{y^2}{xz+yz}+\frac{z}{x+z}+1\)
\(=\frac{1}{\frac{y^2}{xz}+\frac{y}{x}}+\frac{1}{\frac{xz}{y^2}+\frac{z}{y}}+\frac{1}{\frac{x}{z}+1}+1\)
Đặt \((\frac{x}{y}, \frac{y}{z})=(a,b)\Rightarrow ab=\frac{x}{z}\geq 1\) do $x\ge z$
BĐT cần CM trở thành:
\(\frac{1}{\frac{b}{a}+\frac{1}{a}}+\frac{1}{\frac{a}{b}+\frac{1}{b}}+\frac{1}{ab+1}+1\geq \frac{5}{2}\)
\(\Leftrightarrow \frac{a}{b+1}+\frac{b}{a+1}+\frac{1}{ab+1}+1\geq \frac{5}{2}\)
\(\Leftrightarrow \frac{a+b+1}{b+1}+\frac{b+a+1}{a+1}+\frac{1}{ab+1}\geq \frac{7}{2}(*)\)
Đăt \(P=\frac{a+b+1}{b+1}+\frac{b+a+1}{a+1}+\frac{1}{ab+1}\). Áp dụng BĐT Cauchy-Schwarz và AM-GM ta có:
\(P\geq (a+b+1).\frac{4}{b+1+a+1}+\frac{1}{(\frac{a+b}{2})^2+1}=\frac{4(a+b+1)}{a+b+2}+\frac{4}{(a+b)^2+4}(1)\)
Đặt \(t=a+b\). Theo BĐT AM-GM \(t=a+b\geq 2\sqrt{ab}\geq 2\sqrt{1}=2\)
Xét hiệu:
\(\frac{4(a+b+1)}{a+b+2}+\frac{4}{(a+b)^2+4}-\frac{7}{2}=\frac{4(t+1)}{t+2}+\frac{4}{t^2+4}-\frac{7}{2}\)
\(=\frac{t^3-6t^2+12t-8}{2(t+2)(t^2+4)}=\frac{(t-2)^3}{2(t+2)(t^2+4)}\geq 0, \forall t\geq 2\)
\(\Rightarrow \frac{4(a+b+1)}{a+b+2}+\frac{4}{(a+b)^2+4}\geq \frac{7}{2}(2)\)
Từ \((1);(2)\Rightarrow P\geq \frac{7}{2}\). BĐT $(*)$ đúng, ta có đpcm.
Dấu "=" xảy ra khi $x=y=z$
\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)
\(\Rightarrow xyz\le1\)
\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)
Ta co:
\(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x\sqrt[3]{x}}{\sqrt[3]{xyz}}+\frac{y\sqrt[3]{y}}{\sqrt[3]{xyz}}+\frac{z\sqrt[3]{z}}{\sqrt[3]{xyz}}\)
\(\ge x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\)
\(\Rightarrow3A\ge3\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\ge\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)
\(\ge\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow A\ge xy+yz+zx\)
Áp dụng BĐT Cauchy - Schwarz, ta có: \(3\left(x^2+y^2+z^2\right)=\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3=x^2+y^2+z^2\)(Do \(x^2+y^2+z^2=3\))
Ta có: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}=\frac{x}{\sqrt[3]{yz.1}}+\frac{y}{\sqrt[3]{zx.1}}+\frac{z}{\sqrt[3]{xy.1}}\)
\(\ge\frac{x}{\frac{y+z+1}{3}}+\frac{y}{\frac{z+x+1}{3}}+\frac{z}{\frac{x+y+1}{3}}\)\(=\frac{3x}{y+z+1}+\frac{3y}{z+x+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+zx+x}+\frac{3y^2}{yz+xy+y}+\frac{3z^2}{zx+yz+z}\)\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+\left(x+y+z\right)}\)(Theo BĐT Cauchy - Schwarz dạng Engle)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\)
\(\ge xy+yz+zx\)
Đẳng thức xảy ra khi x = y = z = 1
Trước hết ta chứng minh bổ đề sau đây: \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{9\left(a^2+b^2+c^2\right)}{\left(a+b+c\right)^2}\). Đặt P = VT - VP.
(đây là phân tích của một người khác, không phải của em)
Do đó \(VT=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\frac{9\left(x^2+y^2+z^2\right)}{\left(x+y+z\right)^2}=\frac{27}{\sqrt{\left(x+y+z\right)^2.\left(x+y+z\right)^2}}\)
\(\ge\frac{27}{\sqrt{3\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2}}=\frac{9}{x+y+z}\)
Đẳng thức xảy ra khi x = y = z = 1
P/s: Em không chắc lắm!
Theo giả thiết: \(x^2+y^2+z^2=3\Rightarrow2\left(xy+yz+zx\right)=\left(x+y+z\right)^2-3\)
Theo BĐT Bunyakovsky dạng phân thức, ta có:
\(VT=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{x^2}{xy}+\frac{y^2}{yz}+\frac{z^2}{zx}\)\(\ge\frac{\left(x+y+z\right)^2}{xy+yz+zx}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3}\)
Đến đây, ta cần chỉ ra rằng \(\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3}\ge\frac{9}{x+y+z}\)(*)
Ta có: \(xy+yz+zx>0\Leftrightarrow\left(x+y+z\right)^2\ge x^2+y^2+z^2=3\)
\(\Rightarrow x+y+z>\sqrt{3}\)
Đặt \(x+y+z=t>\sqrt{3}\). Khi đó (*) trở thành \(\frac{2t^2}{t^2-3}\ge\frac{9}{t}\Leftrightarrow\frac{\left(t-3\right)^2\left(2t+3\right)}{t\left(t^2-3\right)}\ge0\)(đúng với mọi \(t>\sqrt{3}\))
Đẳng thức xảy ra khi \(t=3\)hay x = y = z = 1