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\(P=1+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{z^2}{xy}\)
vì \(x^2+y^2=z^2\Rightarrow z=\sqrt{x^2+y^2}\)
Áp dụng BĐT bunyakovsky:
\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
\(\Rightarrow z=\sqrt{x^2+y^2}\ge\sqrt{\dfrac{\left(x+y\right)^2}{2}}=\dfrac{\sqrt{2}\left(x+y\right)}{2}\)
do đó \(P\ge1+\dfrac{\sqrt{2}\left(x+y\right)}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{x^2+y^2}{xy}\)
Áp dụng BĐT cauchy:\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)và\(x^2+y^2\ge2xy\)
\(P\ge1+\dfrac{\sqrt{2}\left(x+y\right)}{2}.\dfrac{4}{x+y}+\dfrac{2xy}{xy}=3+2\sqrt{2}\)
dấu = xảy ra khi \(x=y=\dfrac{\sqrt{2}z}{2}\)
a)\(VT=y^2-2y+3=\left(y-1\right)^2+2\ge2\)
\(VP=\dfrac{6}{x^2+2x+4}=\dfrac{6}{\left(x+1\right)^2+3}\le\dfrac{6}{3}=2\)
Dấu "=" xảy ra khi: \(y=1;x=-1\)
b) Áp dụng bất đẳng thức AM-GM:
\(\sqrt{x-a}\le\dfrac{x-a+1}{2}\)
\(\sqrt{y-b}\le\dfrac{y-b+1}{2}\)
\(\sqrt{z-c}\le\dfrac{z-c+1}{2}\)
Cộng theo vế:
\(VT\le\dfrac{x-a+1+y-b+1+z-c+1}{2}=\dfrac{x+y+z}{2}=VP\)
Dấu "=" xảy ra khi: \(x=y=z=2\)
Ta có BĐT:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(\Leftrightarrow6\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)+2016\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow7.\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le2016\)
Xét \(P=\frac{1}{\sqrt{3\left(2x^2+y^2\right)}}+\frac{1}{\sqrt{3\left(2y^2+z^2\right)}}+\frac{1}{\sqrt{3\left(2z^2+x^2\right)}}\)
\(P^2=\left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2x^2+y^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2y^2+z^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2z^2+x^2}}\right)^2\)
Áp dụng BĐT Bunhiacopxki ta có:
\(P^2\le\left(\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2\right)\left(\left(\frac{1}{\sqrt{2x^2+y^2}}\right)^2+\left(\frac{1}{\sqrt{2y^2+z^2}}\right)^2+\left(\frac{1}{\sqrt{2z^2+x^2}}\right)^2\right)\)
\(\Leftrightarrow P^2\le\frac{1}{2x^2+y^2}+\frac{1}{2y^2+z^2}+\frac{1}{2z^2+x^2}\)
Mặt khác ta có:
\(\frac{1}{2x^2+y^2}=\frac{1}{x^2+x^2+y^2}\le\frac{1}{9}\left(\frac{1}{x^2}+\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\frac{1}{2y^2+z^2}\le\frac{1}{9}\left(\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
\(\frac{1}{2z^2+x^2}\le\frac{1}{9}\left(\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}\right)\)
\(\Rightarrow P^2\le\frac{1}{3}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le\frac{1}{3}.2016=672\)
\(\Rightarrow P\le4\sqrt{42}\)
Dấu '=' xảy ra khi \(x=y=z=\sqrt{\frac{1}{672}}\)
Dự đoán dấu = xảy ra khi x=y=\(\dfrac{z}{2}\)
ta có: \(VT=3+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+\dfrac{y^2}{z^2}+\dfrac{z^2}{y^2}+\dfrac{x^2}{z^2}+\dfrac{z^2}{x^2}\)
\(=3+\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)+\left(\dfrac{y^2}{z^2}+\dfrac{x^2}{z^2}\right)+\left(\dfrac{z^2}{y^2}+\dfrac{z^2}{x^2}\right)\)
Áp dụng BĐT AM-GM: \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\ge2\)
Áp dụng BĐT bunyakovsky:\(\dfrac{y^2}{z^2}+\dfrac{x^2}{z^2}\ge\dfrac{1}{2}\left(\dfrac{y}{z}+\dfrac{x}{z}\right)^2=\dfrac{1}{2}.\dfrac{\left(x+y\right)^2}{z^2}\)
\(\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}\ge\dfrac{1}{2}\left(\dfrac{z}{x}+\dfrac{z}{y}\right)^2\ge\dfrac{1}{2}\left(\dfrac{4z}{x+y}\right)^2=\dfrac{8z^2}{\left(x+y\right)^2}\)(AM-GM)
do đó \(VT\ge5+\dfrac{1}{2}\dfrac{\left(x+y\right)^2}{z^2}+\dfrac{8z^2}{\left(x+y\right)^2}\)
Đặt \(\dfrac{z}{x+y}=a\)(a>0)thì \(a\ge1\)do \(z\ge x+y\)
\(VT\ge8a^2+\dfrac{1}{2a^2}+5=\dfrac{a^2}{2}+\dfrac{1}{2a^2}+\dfrac{15}{2}a^2+5\ge\dfrac{a^2}{2}+\dfrac{1}{2a^2}+\dfrac{25}{2}\)
Áp dụng BĐT AM-GM: \(\dfrac{a^2}{2}+\dfrac{1}{2a^2}\ge2\sqrt{\dfrac{a^2}{4a^2}}=1\)
do đó \(VT\ge1+\dfrac{25}{2}=\dfrac{27}{2}\)(đpcm)
Dấu = xảy ra khi a=1 hay \(x=y=\dfrac{z}{2}\)
Lời giải:
Từ \(xy+yz+xz=xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(\left(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\right)=(a,b,c)\Rightarrow a+b+c=1\)
Bài toán tương đương với việc chứng minh:
\(\frac{c^3}{(a+1)(b+1)}+\frac{a^3}{(b+1)(c+1)}+\frac{b^3}{(a+1)(c+1)}\geq \frac{1}{16}\)
Thật vậy, áp dụng BĐT AM-GM ta có:
\(\frac{c^3}{(a+1)(b+1)}+\frac{a+1}{64}+\frac{b+1}{64}\geq 3\sqrt[3]{\frac{c^3}{64^2}}=\frac{3c}{16}\)
Tương tự:
\(\frac{a^3}{(b+1)(c+1)}+\frac{b+1}{64}+\frac{c+1}{64}\geq \frac{3a}{16}\)
\(\frac{b^3}{(c+1)(a+1)}+\frac{c+1}{64}+\frac{a+1}{64}\geq \frac{3c}{16}\)
Cộng các BĐT thu được ở trên:
\(\Rightarrow \text{VT}+\frac{(a+b+c)+3}{32}\geq \frac{3}{16}(a+b+c)\)
\(\Leftrightarrow \text{VT}+\frac{1}{8}\geq \frac{3}{16}\Rightarrow \text{VT}\geq \frac{1}{16}\)
Ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=3\)
from giả thiết => x+y+z=xyz
biến đổi như sau:\(\dfrac{x}{\sqrt{yz\left(1+x^2\right)}}=\dfrac{x}{\sqrt{yz+x^2yz}}=\dfrac{x}{\sqrt{yz+x\left(x+y+z\right)}}=\dfrac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}\)
=\(\sqrt{\dfrac{x^2}{\left(x+y\right)\left(x+z\right)}}\le\dfrac{1}{2}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\)
\(\dfrac{1}{x^2\left(y-z\right)}=-\dfrac{3}{5}\Rightarrow x^2=-\dfrac{5}{3\left(y-z\right)}\)
\(\dfrac{1}{y^2\left(z-x\right)}=\dfrac{1}{3}\Rightarrow y^2=\dfrac{3}{\left(z-x\right)}\)
\(\dfrac{1}{z^2\left(x-y\right)}=3\Rightarrow z^2=\dfrac{1}{3\left(x-y\right)}\)
\(A=x^2.y^2.z^2=-\dfrac{5}{3\left(y-z\right)}.\dfrac{3}{z-x}.\dfrac{1}{3\left(x-y\right)}=\)
\(=-\dfrac{5}{3}.\dfrac{1}{\left(y-z\right)\left(z-x\right)\left(x-y\right)}=\)