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+\(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)
+\(3+2\left(xy+yz+zx\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)=\left(x+y+z\right)^2\le9\)
\(\Rightarrow B=\frac{1}{1+\sqrt{3+2\left(xy+yz+zx\right)}}\ge\frac{1}{1+3}=\frac{1}{4}\)
+\(A=\frac{x^2}{y+2z}+\frac{y^2}{z+2x}+\frac{z^2}{x+2y}=\frac{x^4}{x^2y+2zx^2}+\frac{y^4}{y^2z+2xy^2}+\frac{z^4}{z^2x+2yz^2}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2y+y^2z+z^2x+2\left(xy^2+yz^2+zx^2\right)}\)
Áp dụng bđt Bunhiacopxki
\(x^2y+y^2z+z^2x=x.xy+y.yz+z.zx\le\sqrt{x^2+y^2+z^2}.\sqrt{x^2y^2+y^2z^2+z^2x^2}\)
\(\le\sqrt{x^2+y^2+z^2}.\sqrt{\frac{\left(x^2+y^2+z^2\right)^2}{3}}=3\)
(áp dụng \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\))
Tương tự: \(xy^2+yz^2+zx^2\le3\)
\(\Rightarrow B\ge\frac{3^2}{3+2.3}=1\)
\(VT=A+B\ge1+\frac{1}{4}=\frac{5}{4}=VP\)
Cho a, b, c mà bắt chứng minh x, y, z nên ko chứng minh đc là đúng òi:)
\(VT-VP=\Sigma_{cyc}\frac{\left(x-y\right)^4}{4xy\left(x^2+y^2\right)}\ge0\)
\(ab+bc+ca\le a^2+b^2+c^2\le\frac{\left(a+b+c\right)^2}{3}\) ( bđt phụ + Cauchy-Schwarz dạng Engel )
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
CM bđt phụ : \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Leftrightarrow\)\(2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)
\(\Leftrightarrow\)\(2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)
\(\Leftrightarrow\)\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\ge0\)
\(\Leftrightarrow\)\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) ( luôn đúng )
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z\)
Chúc bạn học tốt ~
Áp dụng BĐT AM-GM ta có:
\(VT=\dfrac{2x^2+y^2+z^2}{4-yz}+\dfrac{2y^2+z^2+x^2}{4-xz}+\dfrac{2z^2+x^2+y^2}{4-xy}\)
\(\ge\dfrac{4x\sqrt{yz}}{4-yz}+\dfrac{4y\sqrt{xz}}{4-xz}+\dfrac{4z\sqrt{xy}}{4-xy}\)
Cần chứng minh: \(\dfrac{4x\sqrt{yz}}{4-yz}+\dfrac{4y\sqrt{xz}}{4-xz}+\dfrac{4z\sqrt{xy}}{4-xy}\ge4xyz\)
\(\Leftrightarrow\dfrac{\sqrt{yz}}{yz\left(4-yz\right)}+\dfrac{\sqrt{xz}}{xz\left(4-xz\right)}+\dfrac{\sqrt{xy}}{xy\left(4-xy\right)}\ge1\)
Mà theo BĐT Cauchy-SChwarz ta có:
\(\left(x+y+z\right)^2\ge\left(1+1+1\right)\left(xy+yz+xz\right)\ge\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)^2\)
\(\Leftrightarrow3\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)
Đăt \(\left(\sqrt{xy};\sqrt{yz};\sqrt{xz}\right)\rightarrow\left(a;b;c\right)\Rightarrow\left\{{}\begin{matrix}a,b,c>0\\a+b+c\le3\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{a}{a^2\left(4-a^2\right)}+\dfrac{b}{b^2\left(4-b^2\right)}+\dfrac{c}{c\left(4-c^2\right)}\ge1\left(\odot\right)\)
Ta có BĐT phụ: \(\dfrac{a}{a^2\left(4-a^2\right)}\le-\dfrac{1}{9}a+\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{\left(a-1\right)^2\left(a^2-2a-9\right)}{9a\left(a-2\right)\left(a+2\right)}\le0\forall0< a\le1\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT_{\left(\odot\right)}\ge\dfrac{-\left(a+b+c\right)}{9}+\dfrac{4}{9}\cdot3\ge\dfrac{-3}{9}+\dfrac{12}{9}=1=VP_{\left(\odot\right)}\)
Dấu "=" \(\Leftrightarrow x=y=z=1\)
a) Giả sử \(x^2-xy+y^2\ge\frac{1}{3}\left(x^2+xy+y^2\right)\)
\(\Leftrightarrow3\left(x^2-xy+y^2\right)\ge\frac{1}{3}.3\left(x^2+xy+y^2\right)\)
\(\Leftrightarrow3\left(x^2-xy+y^2\right)\ge x^2+xy+y^2\)
\(\Leftrightarrow3x^2-3xy+3y^2-x^2-xy-y^2\ge0\)
\(\Leftrightarrow2x^2-4xy+2y^2\ge0\)
\(\Leftrightarrow2\left(x^2-2xy+y^2\right)\ge0\)
\(\Leftrightarrow2\left(x-y\right)^2\ge0\)(luôn đúng với mọi \(x,y\in R\)).
Dấu bằng xảy ra\(\Leftrightarrow x=y\).
Vậy \(x^2-xy+y^2\ge\frac{1}{3}\left(x^2+xy+y^2\right)\)với \(x,y\in R\).
Đặt \(A=\frac{x\sqrt{x}}{x+\sqrt{xy}+y}+\frac{y\sqrt{y}}{y+\sqrt{yz}+z}+\frac{z\sqrt{z}}{z+\sqrt{zx}+x}\left(x,y,z>0\right)\)
Và đặt \(B=\frac{y\sqrt{y}}{x+\sqrt{xy}+y}+\frac{z\sqrt{z}}{y+\sqrt{yz}+z}+\frac{x\sqrt{x}}{z+\sqrt{zx}+x}\left(x,y,z>0\right)\)
Đặt \(\sqrt{x}=m,\sqrt{y}=n,\sqrt{z}=p\left(m,n,p>0\right)\)thì theo đề bài : \(m+n+p=2\)
Lúc đó:
\(A=\frac{m^2.m}{m^2+mn+n^2}+\frac{n^2.n}{n^2+np+p^2}+\frac{p^2.p}{p^2+pm+m^2}\)
\(A=\frac{m^3}{m^2+mn+n^2}+\frac{n^3}{n^2+np+p^2}+\frac{p^3}{p^2+pm+m^2}\)
Và \(B=\frac{n^3}{m^2+mn+n^2}+\frac{p^3}{n^2+np+p^2}+\frac{m^3}{p^2+pm+m^2}\)
Xét hiệu \(A-B=\frac{m^3-n^3}{m^2+mn+n^2}+\frac{n^3-p^3}{n^2+np+p^2}+\frac{p^3-m^3}{p^2+pm+m^2}\)
\(\Leftrightarrow A-B=\frac{\left(m-n\right)\left(m^2+mn+n^2\right)}{m^2+mn+n^2}+\frac{\left(n-p\right)\left(n^2+np+p^2\right)}{n^2+np+p^2}\)\(+\frac{\left(p-m\right)\left(p^2+pm+m^2\right)}{p^2+pm+m^2}\)
\(\Leftrightarrow A-B=\left(m-n\right)+\left(n-p\right)+\left(p-m\right)\)
\(\Leftrightarrow A-B=m-n+n-p+p-m=0\)
\(\Leftrightarrow A=B\)
Xét \(A+B=\frac{m^3+n^3}{m^2+mn+n^2}+\frac{n^3+p^3}{n^2+np+p^2}+\frac{p^3+m^3}{p^2+pm+m^2}\)
\(\Leftrightarrow A+A=2A=\frac{\left(m+n\right)\left(m^2-mn+n^2\right)}{m^2+m+n^2}+\frac{\left(n+p\right)\left(n^2-np+p^2\right)}{n^2+np+p^2}\)\(\frac{\left(p+m\right)\left(p^2-pm+m^2\right)}{p^2+pm+m^2}\)
Theo câu a), ta có \(x^2-xy+y^2\ge\frac{1}{3}\left(x^2+xy+y^2\right)\)với \(x,y\in R\)
\(\Leftrightarrow\frac{x^2-xy+y^2}{x^2+xy+y^2}\ge\frac{1}{3}\left(1\right)\)
Dấu bằng xảy ra \(\Leftrightarrow x=y\)
Áp dụng bất đẳng thức (1) (với \(m,n>0\)), ta được:
\(\frac{m^2-mn+n^2}{m^2+mn+n^2}\ge\frac{1}{3}\)
\(\Leftrightarrow\frac{\left(m+n\right)\left(m^2-mn+n^2\right)}{m^2+mn+n^2}\ge\frac{m+n}{3}\left(2\right)\)
Dấu bằng xảy ra \(\Leftrightarrow m=n>0\)
Chứng minh tương tự, ta được:
\(\frac{\left(n+p\right)\left(n^2-np+p^2\right)}{n^2+np+p^2}\ge\frac{n+p}{3}\left(3\right)\)
Dấu bằng xảy ra\(\Leftrightarrow n=p>0\)
\(\frac{\left(p+m\right)\left(p^2-pm+m^2\right)}{p^2+pm+m^2}\ge\frac{p+m}{2}\left(4\right)\)
Dấu bằng xảy ra\(\Leftrightarrow p=m>0\)
Từ \(\left(2\right),\left(3\right),\left(4\right)\), ta được:
\(\frac{\left(m+n\right)\left(m^2-mn+n^2\right)}{m^2+mn+n^2}+\frac{\left(n+p\right)\left(n^2-np+p^2\right)}{n^2+np+p^2}\)\(+\frac{\left(p+m\right)\left(p^2-pm+m^2\right)}{p^2-pm+m^2}\ge\frac{m+n}{3}+\frac{n+p}{3}+\frac{p+m}{3}\)
\(\Leftrightarrow2A\ge\frac{m+n+n+p+p+m}{3}\)
\(\Leftrightarrow2A\ge\frac{2\left(m+n+p\right)}{3}\)
\(\Leftrightarrow A\ge\frac{m+n+p}{3}\)
\(\Leftrightarrow A\ge\frac{2}{3}\)(vì \(m+n+p=2\)) (điều phải chứng minh).
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}m=n=p>0\\m+n+p=2\end{cases}}\Leftrightarrow m=n=p=\frac{2}{3}\)\(\Leftrightarrow\sqrt{x}=\sqrt{y}=\sqrt{z}=\frac{2}{3}\Leftrightarrow x=y=z=\frac{4}{9}\)
Vậy nếu \(x,y,z>0\) và \(\sqrt{x}+\sqrt{y}+\sqrt{z}=2\)thì: \(\frac{x\sqrt{x}}{x+\sqrt{xy}+y}+\frac{y\sqrt{y}}{y+\sqrt{yz}+z}+\frac{z\sqrt{z}}{z+\sqrt{zx}+x}\ge\frac{2}{3}\).
Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)
Khi đó BĐT <=>
\(\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+z\right)\left(x+z\right)}+\frac{1}{\left(x+y\right)\left(y+z\right)}\ge\frac{2}{3}\left(\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}+...\right)\)
<=> \(\frac{x+y+z}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\frac{x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}}{\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}\right)^3\)
<=>\(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge\frac{1}{3}\left(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}\right)^3\)
<=> \(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\right)^3\)(1)
Xét \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)
<=> \(9\left[xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\right]\ge8\left(xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+3xyz\right)\)
<=> \(xy\left(y+x\right)+yz\left(y+z\right)+xz\left(x+z\right)\ge6xyz\)
<=> \(x\left(y-z\right)^2+z\left(x-y\right)^2+y\left(x-z\right)^2\ge0\)luôn đúng
Khi đó (1) <=>
\(\left(x+y+z\right).\frac{2\sqrt{2}}{3}.\sqrt{x+y+z}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+....\right)^3\)
<=> \(\sqrt{2\left(x+y+z\right)}\ge\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\)
Áp dụng buniacopxki cho vế phải ta có
\(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\le\sqrt{\left(x+y+z\right)\left(3-xy-yz-xz\right)}\)
\(=\sqrt{2\left(x+y+z\right)}\)
=> BĐT được CM
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
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
vì có 1 chút nhầm lẫn nên giờ mk mới ra mong bạn thứ lỗi
bài 1
\(\Leftrightarrow\frac{4a^4}{2a^3+2a^2b^2}+\frac{4b^4}{2b^3+2c^2b^2}+\frac{4c^4}{2c^3+2a^2c^2}\)
\(\ge\frac{\left(2a^2+2b^2+2c^2\right)^2}{2a^3+2b^3+2c^3+2a^2b^2+2c^2b^2+2a^2c^2}\)
\(\ge\frac{36}{a^4+a^2+b^4+b^2+c^4+c^2+2a^2b^2+2c^2b^2+2a^2c^2}\)
\(=\frac{36}{\left(a^2+b^2+c^2\right)^2+a^2+b^2+c^2}=3\ge a+b+c\)
Dấu bằng xảy ra khi \(a=b=c=1\)
Bài 2 là chuyên Bình Thuận, 2016-2017
Áp dụng bất đẳng thức Cauchy – Schwarz, ta có:
\(\frac{xy}{x^2+yz+zx}\le\frac{xy\left(y^2+yz+zx\right)}{\left(x^2+yz+zx\right)\left(y^2+yz+zx\right)}\le\frac{xy\left(y^2+yz+zx\right)}{\left(xy+yz+zx\right)^2}\)
Tương tự: \(\frac{yz}{y^2+zx+xy}\le\frac{xy\left(z^2+zx+xy\right)}{\left(xy+yz+zx\right)^2}\);\(\frac{zx}{z^2+xy+yz}\le\frac{zx\left(x^2+xy+yz\right)}{\left(xy+yz+zx\right)^2}\)
Cộng từng vế của 3 BĐT trên. ta được:
\(VT\le\frac{\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)}{\left(xy+yz+zx\right)^2}=\frac{x^2+y^2+z^2}{xy+yz+zx}\)
Đẳng thức xảy ra khi x = y = z
hùi nãy mem nào k sai cho t T_T t buồn
\(VT\ge6\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-2\left(xy+yz+zx\right)+2.\frac{9}{4\left(x+y+z\right)}\)
\(=6\left(x+y+z\right)^2-2.\frac{\left(x+y+z\right)^2}{3}+\frac{9}{2\left(x+y+z\right)}=6.\left(\frac{3}{4}\right)^2-2.\frac{\left(\frac{3}{4}\right)^2}{3}+\frac{9}{2.\frac{3}{4}}\)
\(=\frac{27}{8}-\frac{3}{8}+6=9\)
\(\Rightarrow\)\(VT\ge9\) ( đpcm )
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{4}\)
Chúc bạn học tốt ~
Áp dụng bất đẳng thức AM - GM cho các bộ bốn số không âm, ta được: \(LHS=\frac{2x^2+y^2+z^2}{4-yz}+\frac{2y^2+z^2+x^2}{4-zx}+\frac{2z^2+x^2+y^2}{4-xy}\)\(=\frac{x^2+x^2+y^2+z^2}{4-yz}+\frac{y^2+y^2+z^2+x^2}{4-zx}+\frac{z^2+z^2+x^2+y^2}{4-xy}\)\(\ge\frac{4x\sqrt{yz}}{4-yz}+\frac{4y\sqrt{zx}}{4-zx}+\frac{4z\sqrt{xy}}{4-xy}\)
Như vậy, ta cần chứng minh: \(\frac{4x\sqrt{yz}}{4-yz}+\frac{4y\sqrt{zx}}{4-zx}+\frac{4z\sqrt{xy}}{4-xy}\ge4xyz\)\(\Leftrightarrow\frac{\sqrt{yz}}{yz\left(4-yz\right)}+\frac{\sqrt{zx}}{zx\left(4-zx\right)}+\frac{\sqrt{xy}}{xy\left(4-xy\right)}\ge1\)
Theo bất đẳng thức Cauchy-Schwarz, ta có: \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\ge\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2\)
\(\Rightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\le3\)
Đặt \(\left(\sqrt{xy};\sqrt{yz};\sqrt{zx}\right)\rightarrow\left(a;b;c\right)\). Khi đó \(\hept{\begin{cases}a,b,c>0\\a+b+c\le3\end{cases}}\)
và ta cần chứng minh \(\frac{a}{a^2\left(4-a^2\right)}+\frac{b}{b^2\left(4-b^2\right)}+\frac{c}{c^2\left(4-c^2\right)}\ge1\)
Xét BĐT phụ: \(\frac{x}{x^2\left(4-x^2\right)}\ge-\frac{1}{9}x+\frac{4}{9}\left(0< x\le1\right)\)(*)
Ta có: (*)\(\Leftrightarrow\frac{\left(x-1\right)^2\left(x^2-2x-9\right)}{9x\left(x-2\right)\left(x+2\right)}\ge0\)(Đúng với mọi \(x\in(0;1]\))
Áp dụng, ta được: \(\frac{a}{a^2\left(4-a^2\right)}+\frac{b}{b^2\left(4-b^2\right)}+\frac{c}{c^2\left(4-c^2\right)}\ge-\frac{1}{9}\left(a+b+c\right)+\frac{4}{9}.3\)
\(\ge-\frac{1}{9}.3+\frac{4}{3}=1\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi a = b = c = 1
1. Chứng minh với mọi số thực a, b, c ta có 2a2+b2+c2\(\ge\)2a(b+c)
Chứng minh:
Ta có 2a2+b2+c2=(a2+b2)+(a2+c2)
Áp dụng bđt cauchy ta có
(a2+b2)+(a2+c2)\(\ge\)2ab+2ac=2a(b+c)