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Đoạn cuối của cô Nguyễn Linh Chi em có 1 cách biến đổi tương đương cũng khá ngắn gọn ạ
\(RHS\ge2\cdot\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)
Theo đánh giá của cô Nguyễn Linh Chi thì \(xy+yz+zx\ge x+y+z\ge3\)
Ta cần chứng minh:\(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\ge\frac{1}{2}\)
Thật vậy,BĐT tương đương với:
\(2\left(x+y+z\right)^2\ge x^2+y^2+z^2-x-y-z+18\)
\(\Leftrightarrow\left(x+y+z\right)^2+x+y+z-12\ge0\)
\(\Leftrightarrow\left(x+y+z+4\right)\left(x+y+z-3\right)\ge0\) ( luôn đúng với \(x+y+z\ge3\) )
=> đpcm
Áp dụng: \(AB\le\frac{\left(A+B\right)^2}{4}\)với mọi A, B
Ta có:
\(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\le\frac{\left(x+2+x^2-2x+4\right)^2}{4}\)
=> \(\sqrt{x^3+8}\le\frac{x^2-x+6}{2}\)
=> \(\frac{x^2}{\sqrt{x^3+8}}\ge\frac{2x^2}{x^2-x+6}\)
Tương tự
=> \(\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}\)
\(\ge\frac{2x^2}{x^2-x+6}+\frac{2y^2}{y^2-y+6}+\frac{2z^2}{z^2-z+6}\)
\(=2\left(\frac{x^2}{x^2-x+6}+\frac{y^2}{y^2-y+6}+\frac{z^2}{z^2-z+6}\right)\)
\(\ge2\frac{\left(x+y+z\right)^2}{x^2-x+6+y^2-y+6+z^2-z+6}\)
\(=2\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)(1)
Ta có: \(x+y+z\le xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\) với mọi x, y, z
=> \(\left(x+y+z\right)^2-3\left(x+y+z\right)\ge0\)
=> \(\left(x+y+z\right)\left(x+y+z-3\right)\ge0\)
=> \(x+y+z\ge3\)với mọi x, y, z dương
Và \(x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+zx\right)\le\left(x+y+z\right)^2-2\left(x+y+z\right)\)
Do đó: \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)
\(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3\left(x+y+z\right)+18}\)
Đặt: x + y + z = t ( t\(\ge3\))
Xét hiệu: \(\frac{t^2}{t^2-3t+18}-\frac{1}{2}=\frac{t^2+3t-18}{t^2-3t+18}=\frac{\left(t-3\right)\left(t+6\right)}{\left(t-\frac{3}{2}\right)^2+\frac{63}{4}}\ge0\)với mọi t \(\ge3\)
Do đó: \(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3\left(x+y+z\right)+18}\ge\frac{1}{2}\)(2)
Từ (1); (2)
=> \(\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}\ge2.\frac{1}{2}=1\)
Dấu "=" xảy ra <=> x= y = z = 1
Đặt \(\sqrt{x};\sqrt{y};\sqrt{z}\rightarrow a,b,c\), ta có : \(a+b+c=1\)
Tìm min của \(A=\frac{ab}{\sqrt{5a^2+32ab+12b^2}}+\frac{bc}{\sqrt{5b^2+32bc+12c^2}}+\frac{ca}{\sqrt{5c^2+32ca+12a^2}}\)
đến đây thấy giống giống bài bất của HN năm nào ấy nhỉ ?
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 ..............
\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)
\(\Rightarrow xyz\le1\)
\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)
Ta co:
\(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x\sqrt[3]{x}}{\sqrt[3]{xyz}}+\frac{y\sqrt[3]{y}}{\sqrt[3]{xyz}}+\frac{z\sqrt[3]{z}}{\sqrt[3]{xyz}}\)
\(\ge x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\)
\(\Rightarrow3A\ge3\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\ge\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)
\(\ge\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow A\ge xy+yz+zx\)
Áp dụng BĐT Cauchy - Schwarz, ta có: \(3\left(x^2+y^2+z^2\right)=\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3=x^2+y^2+z^2\)(Do \(x^2+y^2+z^2=3\))
Ta có: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}=\frac{x}{\sqrt[3]{yz.1}}+\frac{y}{\sqrt[3]{zx.1}}+\frac{z}{\sqrt[3]{xy.1}}\)
\(\ge\frac{x}{\frac{y+z+1}{3}}+\frac{y}{\frac{z+x+1}{3}}+\frac{z}{\frac{x+y+1}{3}}\)\(=\frac{3x}{y+z+1}+\frac{3y}{z+x+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+zx+x}+\frac{3y^2}{yz+xy+y}+\frac{3z^2}{zx+yz+z}\)\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+\left(x+y+z\right)}\)(Theo BĐT Cauchy - Schwarz dạng Engle)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\)
\(\ge xy+yz+zx\)
Đẳng thức xảy ra khi x = y = z = 1
\(\frac{\left(x+y+z\right)^2}{3}\ge xy+yz+zx\Rightarrow x+y+z\ge3\)
\(P=\frac{x^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}}+\frac{y^2}{\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}}+\frac{z^2}{\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\)
\(\Rightarrow P\ge\frac{\left(x+y+z\right)^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}+\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}+\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\)
\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x+2+x^2-2x+4\right)+\left(y+2+y^2-2y+4\right)+\left(z+2+z^2-2z+4\right)}\)
\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x^2+y^2+z^2\right)-\left(x+y+z\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)-2\left(xy+yz+zx\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)+12}\)
Dự đoán Min P=1 khi x+y+z=3
Đặt \(t=x+y+z\ge3\)
\(\Rightarrow P\ge\frac{2t^2}{t^2-t+12}\Rightarrow P-1\ge\frac{t^2+t-12}{t^2-t+12}=\frac{\left(t-3\right)\left(t+4\right)}{t^2-t+12}\ge0\)
\(\Rightarrow P\ge1\)
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