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![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
solution:
ta có: \(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\Leftrightarrow xyz\le1\)(theo BĐT cauchy cho 3 số )
\(\Rightarrow xy\le\dfrac{1}{z};yz\le\dfrac{1}{x};xz\le\dfrac{1}{y}\)
\(\Rightarrow\dfrac{x}{\sqrt[3]{yz}}\ge\dfrac{x}{\dfrac{1}{\sqrt[3]{x}}}=x\sqrt[3]{x}=\sqrt[3]{x^4}\)
tương tự ta có:\(\dfrac{y}{\sqrt[3]{xz}}\ge\sqrt[3]{y^4};\dfrac{z}{\sqrt[3]{xy}}\ge\sqrt[3]{z^4}\)
cả 2 vế các BĐT đều dương,cộng vế với vế:
\(S=\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}\ge\sqrt[3]{x^4}+\sqrt[3]{y^4}+\sqrt[3]{z^4}\)
Áp dụng BĐT bunyakovsky ta có:
\(\left(\sqrt[3]{x^4}+\sqrt[3]{y^4}+\sqrt[3]{z^4}\right)\left(x^2+y^2+z^2\right)\ge\left(\sqrt[3]{x^8}+\sqrt[3]{y^8}+\sqrt[3]{z^8}\right)^2=\left(x^2+y^2+z^2\right)^2\)
\(\Rightarrow S\ge x^2+y^2+z^2\)
đến đây ta lại có BĐT quen thuộc: \(x^2+y^2+z^2\ge xy+yz+xz\)
\(\Rightarrow S\ge xy+yz+xz\left(đpcm\right)\)
dấu = xảy ra khi và chỉ khi x=y=z mà x2+y2+z2=3 => x=y=z=1
*cách khác : Áp dụng BĐT cauchy - schwarz(bunyakovsky):
\(S=\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}=\dfrac{x^4}{x^3.\dfrac{1}{\sqrt[3]{x}}}+\dfrac{y^4}{y^3.\dfrac{1}{\sqrt[3]{y}}}+\dfrac{z^4}{z^3.\dfrac{1}{\sqrt[3]{z}}}\)
\(S\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2}=x^2+y^2+z^2\ge xy+yz+xz\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\)
\(\Rightarrow\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}\ge\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\)
\(\Rightarrow\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}-\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\ge0\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{2}{\sqrt{xy}}+\dfrac{1}{y}+\dfrac{1}{y}-\dfrac{2}{\sqrt{yz}}+\dfrac{1}{z}+\dfrac{1}{z}-\dfrac{2}{\sqrt{zx}}+\dfrac{1}{x}\ge0\)
\(\Rightarrow\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{y}}\right)^2+\left(\dfrac{1}{\sqrt{y}}-\dfrac{1}{\sqrt{z}}\right)^2+\left(\dfrac{1}{\sqrt{z}}-\dfrac{1}{\sqrt{x}}\right)^2\ge0\) (luôn đúng)
Dấu = xảy ra khi \(x=y=z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)
\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)
\(=\dfrac{x+y+z}{x^2+y^2+z^2-\left(xy+yz+zx\right)+3\left(xy+yz+zx\right)}\)
(vì \(2013=3.671=3\left(xy+yz+zx\right)\))
\(=\dfrac{x+y+z}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}\)
\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)
\(=\dfrac{1}{x+y+z}\)
ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)
\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)
\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))
Vậy ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
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 $x=y=z$
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Dùng nesbit thì ra