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\(ab+bc+ca\le a^2+b^2+c^2\le\frac{\left(a+b+c\right)^2}{3}\) ( 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 ~
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 ~
a.
\(\dfrac{x}{x+\sqrt{3x+yz}}=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\dfrac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}\)
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế:
\(VT\le\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
b.
\(VP=\dfrac{4\left(a+b+c\right)}{2\sqrt{4a\left(a+3b\right)}+2\sqrt{4b\left(b+3c\right)}+2\sqrt{4c\left(c+3a\right)}}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{4a+a+3b+4b+b+3c+4c+c+3a}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{8\left(a+b+c\right)}=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
\(\left(xy+yz+zx\right)^2\ge3xyz\left(x+y+z\right)=9\Rightarrow xy+yz+zx\ge3\)
\(2\left(x^2+y^2\right)-xy\ge\left(x+y\right)^2-\dfrac{1}{4}\left(x+y\right)^2=\dfrac{3}{4}\left(x+y\right)^2\)
Tương tự và nhân vế với vế:
\(VT\ge\dfrac{27}{64}\left[\left(x+y\right)\left(y+z\right)\left(z+x\right)\right]^2\)
Mặt khác ta có:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)
\(\Rightarrow VT\ge\dfrac{27}{64}.\dfrac{64}{81}.3\left(xy+yz+zx\right)^3\ge3^3=27\) (đpcm)
Lời giải:
Từ \(xy+yz+xz=xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \((a,b,c)=\left(\frac{1}{x}; \frac{1}{y}; \frac{1}{z}\right)\Rightarrow a+b+c=1\)
BĐT cần chứng minh trở thành:
\(P=\frac{c^3}{(a+1)(b+1)}+\frac{a^3}{(b+1)(c+1)}+\frac{b^3}{(c+1)(a+1)}\geq \frac{1}{16}(*)\)
Thật vậy, áp dụng BĐT Cauchy ta có:
\(\frac{c^3}{(a+1)(b+1)}+\frac{a+1}{64}+\frac{b+1}{64}\geq 3\sqrt[3]{\frac{c^3}{64^2}}=\frac{3c}{16}\)
\(\frac{a^3}{(b+1)(c+1)}+\frac{b+1}{64}+\frac{c+1}{64}\geq 3\sqrt[3]{\frac{a^3}{64^2}}=\frac{3a}{16}\)
\(\frac{b^3}{(c+1)(a+1)}+\frac{c+1}{64}+\frac{a+1}{64}\geq 3\sqrt[3]{\frac{b^3}{64^2}}=\frac{3b}{16}\)
Cộng theo vế các BĐT trên và rút gọn :
\(\Rightarrow P+\frac{a+b+c+3}{32}\geq \frac{3(a+b+c)}{16}\)
\(\Leftrightarrow P+\frac{4}{32}\geq \frac{3}{16}\Leftrightarrow P\geq \frac{1}{16}\)
Vậy \((*)\) được chứng minh. Bài toán hoàn tất.
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=3\)
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 \(x=y=z=\sqrt{\frac{1}{672}}\)
\(6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)
\(=6\left(x^2+y^2+z^2\right)+12\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)-2\left(xy+yz+xz\right)\)
\(=6\left(x+y+z\right)^2+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{2z+x+y}\right)-2\left(xy+yz+xz\right)\)
\(\ge6\left(x+y+z\right)^2+2.\dfrac{\left(1+1+1\right)^2}{2x+y+z+x+2y+z+2z+x+y}-2\left(xy+yz+xz\right)\)
\(=6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-2\left(xy+yz+xz\right)\)
\(\ge6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-\dfrac{2}{3}\left(x+y+z\right)^2\)
\(=6.\left(\dfrac{3}{4}\right)^2+\dfrac{18}{4.\dfrac{3}{4}}-\dfrac{2}{3}.\left(\dfrac{3}{4}\right)^2=9\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)
a) ab+bc+ca\(\le\dfrac{\left(a+c+b\right)^2}{3}\)
\(\Leftrightarrow3ab+3bc+3ac\le a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Leftrightarrow ab+bc+ac\le a^2+b^2+c^2\)
\(\Leftrightarrow2ab+2bc+2ca\le2a^2+2b^2+2c^2\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng \(\forall a,b,c\)