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\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)
\(=\frac{1}{\sqrt{\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y-z\right)^2+\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z-x\right)^2+\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{1}{\sqrt{\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Thấy cái đề mà thấy khiếp ...
Ta có : \(x^2-xy+y^2=\frac{3}{4}\left(x^2-2xy+y^2\right)+\frac{1}{4}\left(x^2+2xy+y^2\right)\)
\(=\frac{3}{4}\left(x-y\right)^2+\frac{1}{4}\left(x+y\right)^2\ge\frac{1}{4}\left(x+y\right)^2\)
\(\Rightarrow\sqrt{x^2-xy+y^2}\ge\frac{x+y}{2}\)
Tương tự \(\sqrt{y^2-yz+z^2}\ge\frac{y+z}{2}\)
\(\sqrt{z^2-zx+x^2}\ge\frac{x+z}{2}\)
Do đó : \(2S\ge\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{x+z}{x+z+2y}\)
\(\Rightarrow2S+3\ge\left(1+\frac{x+y}{x+y+2z}\right)+\left(1+\frac{y+z}{y+z+2x}\right)+\left(1+\frac{x+z}{x+z+2y}\right)\)
\(=2\left(x+y+z\right)\left(\frac{1}{x+y+2z}+\frac{1}{y+z+2x}+\frac{1}{x+z+2y}\right)\)
\(\ge2\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}\)\(=\frac{9}{2}\)
(Áp dụng bđt Cô-si dạng engel cho 3 số)
\(\Rightarrow2S+3\ge\frac{9}{2}\)
\(\Rightarrow S\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
Vậy ..............
\(P=\frac{\sqrt{1+x^2+y^2}}{xy}+\frac{\sqrt{1+y^2+z^2}}{yz}+\frac{\sqrt{1+z^2+x^2}}{zx}\)
\(\ge\text{Σ}\frac{\sqrt{\frac{\left(1+x+y\right)^2}{3}}}{xy}\text{=}\frac{1+x+y}{xy\sqrt{3}}\)
\(=\frac{\sqrt{3}}{3}\left(\frac{1+x+y}{xy}+\frac{1+y+z}{yz}+\frac{1+z+x}{zx}\right)\)
\(=\frac{\sqrt{3}}{3}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{x}\right)\)
\(=\frac{\sqrt{3}}{3}\left(x+y+z+2xy+2yz+2zx\right)\)\(\ge\frac{\sqrt{3}}{3}\left(3\sqrt[3]{xyz}+2\cdot3\sqrt[3]{x^2y^2z^2}\right)=\frac{\sqrt{3}}{3}\left(3+6\right)=3\sqrt{3}\)
Dấu = xảy ra khi \(x=y=z=1\)
1. Với mọi số thực x;y;z ta có:
\(x^2+y^2+z^2+\dfrac{1}{2}\left(x^2+1\right)+\dfrac{1}{2}\left(y^2+1\right)+\dfrac{1}{2}\left(z^2+1\right)\ge xy+yz+zx+x+y+z\)
\(\Leftrightarrow\dfrac{3}{2}P+\dfrac{3}{2}\ge6\)
\(\Rightarrow P\ge3\)
\(P_{min}=3\) khi \(x=y=z=1\)
1.1
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=a>0\\\dfrac{1}{\sqrt{y}}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+\sqrt{2-b^2}=2\\b+\sqrt{2-a^2}=2\end{matrix}\right.\)
\(\Rightarrow a-b+\sqrt{2-b^2}-\sqrt{2-a^2}=0\)
\(\Leftrightarrow a-b+\dfrac{\left(a-b\right)\left(a+b\right)}{\sqrt{2-b^2}+\sqrt{2-a^2}}=0\)
\(\Leftrightarrow a=b\Leftrightarrow x=y\)
Thay vào pt đầu:
\(a+\sqrt{2-a^2}=2\Rightarrow\sqrt{2-a^2}=2-a\) (\(a\le2\))
\(\Leftrightarrow2-a^2=4-4a+a^2\Leftrightarrow2a^2-4a+2=0\)
\(\Rightarrow a=1\Rightarrow x=y=1\)
2.
\(\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^2-xy+y^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3xy+3y^2=21\\7x^2-7xy+7y^2=21\end{matrix}\right.\)
\(\Rightarrow4x^2-10xy+4y^2=0\)
\(\Leftrightarrow2\left(2x-y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\dfrac{1}{2}x\end{matrix}\right.\)
Thế vào pt đầu
...
\(\frac{\left(x+y+z\right)^2}{3}\ge xy+yz+zx\Rightarrow x+y+z\ge3\)
\(P=\frac{x^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}}+\frac{y^2}{\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}}+\frac{z^2}{\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\)
\(\Rightarrow P\ge\frac{\left(x+y+z\right)^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}+\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}+\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\)
\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x+2+x^2-2x+4\right)+\left(y+2+y^2-2y+4\right)+\left(z+2+z^2-2z+4\right)}\)
\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x^2+y^2+z^2\right)-\left(x+y+z\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)-2\left(xy+yz+zx\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)+12}\)
Dự đoán Min P=1 khi x+y+z=3
Đặt \(t=x+y+z\ge3\)
\(\Rightarrow P\ge\frac{2t^2}{t^2-t+12}\Rightarrow P-1\ge\frac{t^2+t-12}{t^2-t+12}=\frac{\left(t-3\right)\left(t+4\right)}{t^2-t+12}\ge0\)
\(\Rightarrow P\ge1\)
Hi ~! Mình xin slot trước :)
Giải
Dự đoán dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\) khi đó \(P=\frac{3\sqrt{3}}{4}\)
Ta sẽ chứng minh nó là GTNN của \(P\)
Ta có: \(x^2+xy+y^2=\frac{3\left(x+y\right)^2+\left(x-y\right)^2}{4}\ge\frac{3\left(x+y\right)^2}{4}\)
Do đó ta cần chứng minh
\(\frac{x+y}{4yz+1}+\frac{y+z}{4xz+1}+\frac{x+z}{4xy+1}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{x+y}{\left(y+z\right)^2+1}+\frac{y+z}{\left(x+z\right)^2+1}+\frac{x+z}{\left(x+y\right)^2+1}\ge\frac{3}{2}\)
Ta có: \(x+y+z=\frac{3}{2}\Rightarrow2x+2y+2z=3\)
\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(x+z\right)=2\)
Đặt \(\hept{\begin{cases}a=x+y\\b=y+z\\c=z+x\end{cases}}\) thì ta cần chứng minh
\(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\ge\frac{3}{2}\)\(\forall\hept{\begin{cases}a,b,c>0\\a+b+c=3\end{cases}}\)
Lại có: \(\frac{a}{b^2+1}=a-\frac{ab^2}{b^2+1}\ge a-\frac{ab}{2}\)
Tương tự ta cũng có: \(\frac{b}{c^2+1}\ge b-\frac{bc}{2};\frac{c}{a^2+1}\ge c-\frac{ac}{2}\)
Cộng theo vế các BĐT ta có: \(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\ge a-\frac{ab}{2}+b-\frac{bc}{2}+c-\frac{ac}{2}\)
\(=\left(a+b+c\right)-\frac{ab+bc+ca}{2}\ge3-\frac{3}{2}=\frac{3}{2}\)
BĐT đã được c/m vậy ta có \(P\ge\frac{3\sqrt{3}}{4}\Leftrightarrow x=y=z=\frac{1}{2}\)
\(x^2-xy+y^2=\dfrac{1}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2\ge\dfrac{1}{4}\left(x+y\right)^2\)
\(\Rightarrow\sqrt{x^2-xy+y^2}\ge\sqrt{\dfrac{1}{4}\left(x+y\right)^2}=\dfrac{1}{2}\left(x+y\right)\)
Tương tự: \(\sqrt{y^2-yz+z^2}\ge\dfrac{1}{2}\left(y+z\right)\); \(\sqrt{z^2-zx+x^2}\ge\dfrac{1}{2}\left(z+x\right)\)
Cộng vế:
\(Q\ge\dfrac{1}{2}\left(x+y\right)+\dfrac{1}{2}\left(y+z\right)+\dfrac{1}{2}\left(z+x\right)=x+y+z=3\) (đpcm)
ta có
\(4\left(x^2+xy+y^2\right)\ge3\left(x+y\right)^2\Leftrightarrow\left(x-y\right)^2\ge0\) vì thế \(\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}{2}\left(x+y\right)\)
hoàn toàn tương tự ta sẽ có
\(P\ge\frac{\sqrt{3}}{2}\left(x+y\right)+\frac{\sqrt{3}}{2}\left(y+z\right)+\frac{\sqrt{3}}{2}\left(x+z\right)\)
hay
\(P\ge\sqrt{3}\left(x+y+z\right)=3\sqrt{3}\)
dấu bằng xảy ra khi x=y=z=1
Ta có: \(P=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\)\(=\sqrt{\frac{3}{4}\left(x+y\right)^2+\frac{1}{4}\left(x-y\right)^2}+\sqrt{\frac{3}{4}\left(y+z\right)^2+\frac{1}{4}\left(y-z\right)^2}+\sqrt{\frac{3}{4}\left(z+x\right)^2+\frac{1}{4}\left(x-y\right)^2}\)\(\ge\sqrt{\frac{3}{4}\left(x+y\right)^2}+\sqrt{\frac{3}{4}\left(y+z\right)^2}+\sqrt{\frac{3}{4}\left(z+x\right)^2}=\sqrt{3}\left(x+y+z\right)=3\sqrt{3}\)
Đẳng thức xảy ra khi x = y = z = 1