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Bài 1: Theo đề : \(2ab+6bc+2ac=7abc\) \(;a,b,c>0\)
Chia cả 2 vế cho \(abc>0\Rightarrow\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)
Đặt: \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}x,y,z>0\\2z+6x+2y=7\end{cases}}\)
Khi đó: \(M=\frac{4ab}{a+2b}+\frac{9ac}{a+4c}+\frac{4bc}{b+c}=\frac{4}{2x+y}+\frac{9}{4x+z}+\frac{4}{y+z}\)
\(\Rightarrow M=\frac{4}{2x+y}+2x+y+\frac{9}{4x+z}+4x+z+\frac{4}{y+z}+y+z-\left(2x+y+4x+z+y+z\right)\)
\(=\left(\frac{2}{\sqrt{x+2y}}-\sqrt{x+2y}\right)^2+\left(\frac{3}{\sqrt{4x+z}}-\sqrt{4x+z}\right)^2+\left(\frac{2}{\sqrt{y+z}}-\sqrt{y+z}\right)^2+17\ge17\)
Khi: \(\hept{\begin{cases}x=\frac{1}{2}\\y=z=1\end{cases}}\Rightarrow M=17\)
\(Min_M=17\Leftrightarrow a=2;b=1;c=1\)
ミ★๖ۣۜBăηɠ ๖ۣۜBăηɠ ★彡 chém bài khó nhất rồi nên em xin mạn phép chém bài dễ ạ.
2/\(VT=\Sigma_{cyc}\frac{\left(x+y+z\right)^2-x^2}{x\left(x+y+z\right)+yz}=\Sigma_{cyc}\frac{\left(y+z\right)\left(2x+y+z\right)}{\left(x+y\right)\left(x+z\right)}\)
\(\ge\Sigma_{cyc}\frac{\left(y+z\right)\left(2x+y+z\right)}{\frac{\left(2x+y+z\right)^2}{4}}=\Sigma_{cyc}\frac{4\left(y+z\right)}{2x+y+z}=\Sigma_{cyc}\frac{2\left(y+z-2x\right)}{2x+y+z}+6\)
\(=\Sigma_{cyc}\left(\frac{2\left(x+y+z\right)\left(y+z-2x\right)}{2x+y+z}-\frac{3}{2}\left(y+z-2x\right)\right)+6\)
\(=\Sigma_{cyc}\frac{\left(y+z-2x\right)^2}{2\left(2x+y+z\right)}+6\ge6\)
http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%AAn-b%C3%ACnh-thu%E1%BA%ADn-2016-2017/
tiếp tục câu 2,vì máy bị lỗi nên phải tách ra:
Ta có:\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(=\left(x+y+z\right)\left(\left(x+y+z\right)^2-3\left(xy+xz+yz\right)\right).\)
Dó đó:\(x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\left(x+y+z\right)^2-3\left(xy+yz+xz\right)+2010\right)\)
\(=\left(x+y+z\right)^3.\)(2)
TỪ \(\left(1\right),\left(2\right)\)suy ra \(P\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}.\)
Dấu \(=\)xảy ra khi \(x=y=z=\frac{\sqrt{2010}}{3}\)
2)Ta có:
\(x\left(x^2-yz+2010\right)=x\left(x^2+xy+xz+1340\right)>0\)
Tương tự ta có:\(y\left(y^2-xz+2010\right)>0,z\left(z^2-xy+2010\right)>0\)
Áp dụng svac-xơ ta có:
\(P=\frac{x^2}{x\left(x^2-yz+2010\right)}+\frac{y^2}{y\left(y^2-xz+2010\right)}+\frac{z^2}{z\left(z^2-xy+2010\right)}\)
\(\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)}.\)(1)
\(VT=\frac{\left(yz\right)^2}{x^2yz\left(y+z\right)}+\frac{\left(xz\right)^2}{zxy^2\left(x+z\right)}+\frac{\left(xy\right)^2}{xyz^2\left(x+y\right)}\)
\(VT=\frac{2\left(yz\right)^2}{xy+zx}+\frac{2\left(xz\right)^2}{xy+yz}+\frac{2\left(xy\right)^2}{xz+yz}\ge\frac{2\left(yz+xz+xy\right)^2}{2\left(xy+yz+zx\right)}=xy+yz+zx\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{\sqrt[3]{2}}\)
Dự đoán dấu bằng có khi (x,y,z)(x,y,z) là các hoán vị (0;1;1).
Từ đó ta đánh giá làm mất căn:
Ta có:
\(4\sqrt{2}.\sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}}=\frac{8\left(xy+yz+zx\right)}{\sqrt{\left(x^2+y^2+z^2\right).2\left(xy+yz+zx\right)}}\)\(\ge\frac{16\left(xy+yz+zx\right)}{\left(x+y+z\right)^2}\)
Do đó ta chỉ cần có
\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+\frac{16\left(xy+yz+zx\right)}{\left(x+y+z\right)^2}\ge6\)
Không mất tính tổng quát, giả sử \(x\ge y\ge z\) suy ra \(x\ge y>0,z\ge0\)
Khi đó, ta chứng minh BĐT mạnh hơn
\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{16\left(xy+yz+zx\right)}{\left(x+y+z\right)^2}\ge6\)
\(\Leftrightarrow\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}-\frac{8\left(x^2+y^2+z^2\right)}{\left(x+y+z\right)^2}\ge0\)
\(\Leftrightarrow\left(x+y+z\right)^3\left(x+y+2z\right)\ge8\left(x+z\right)\left(y+z\right)\left(x^2+y^2+z^2\right)\)
Hay \(\left(x+y+z\right)^4+z\left(x+y+z\right)^3\ge8z^2\left(x^2+y^2+z^2\right)+8\left(xy+yz+zx\right)\left(x^2+y^2+z^2\right)\)
Theo AM-GM:\(\left(x+y+z\right)^4=\left(x^2+y^2+z^2+2\left(xy+yz+zx\right)\right)^2\ge8\left(xy+yz+zx\right)\left(x^2+y^2+z^2\right)\)
Vậy ta chỉ cần chứng minh \(z\left(x+y+z\right)^3\ge8z^2\left(x^2+y^2+z^2\right)\)
\(BDT\Leftrightarrow\left(x+y+z\right)^3\ge8z\left(x^2+y^2+z^2\right)\)
Ta có:\(\left(x+y+z\right)^3=x^3+y^3+z^3+3x\left(y^2+z^2\right)+3y\left(z^2+x^2\right)+3z\left(x^2+y^2\right)+6xyz\ge x^3+y^3+z^3+3x^2y+3xy^2+5xyz+8z^3+3z\left(x^2+y^2\right)\)
Suy ra \(\left(x+y+z\right)^3-8z\left(x^2+y^2+z^2\right)\ge x^3+y^3+3x^2y+3xy^2+5xyz-5z\left(x^2+y^2\right)\)
\(=x^3+y^3+3x^2y+3xy^2+5z\left(xy-x^2-y^2\right)\ge x^3+y^3+3x^2y+3xy^2+5y\left(xy-x^2-y^2\right)\)
\(\ge x^3+y^3+3x^2y+3xy^2-5y\left(x^2+y^2\right)\)
\(=\left(x^2-y^2+4\right)\left(x-y\right)\ge0\)
BĐT được chứng minh.
Xét: \(\frac{1-x^2}{x+yz}+\frac{1-y^2}{y+xz}+\frac{1-z^2}{z+xy}\)
Thay thế \(x+y+z=1\)
\(\Leftrightarrow\frac{\left(x+y+z\right)^2-x^2}{x\left(x+y+z\right)+yz}+\frac{\left(x+y+z\right)^2-y^2}{y\left(x+y+z\right)+xz}+\frac{\left(x+y+z\right)^2-z^2}{z\left(x+y+z\right)+xy}\)
Áp dụng hằng đẳng thức hiệu 2 bình phương: \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\frac{\left(y+z\right)\left(2x+y+z\right)}{x^2+xy+xz+yz}+\frac{\left(x+z\right)\left(x+2y+z\right)}{xy+y^2+yz+xz}+\frac{\left(x+y\right)\left(x+y+2z\right)}{xz+zy+z^2+xy}\)
\(\Leftrightarrow\frac{\left(y+z\right)\left(2x+y+z\right)}{\left(x+y\right)\left(x+z\right)}+\frac{\left(x+z\right)\left(x+2y+z\right)}{\left(x+y\right)\left(y+z\right)}+\frac{\left(x+y\right)\left(x+y+2z\right)}{\left(x+z\right)\left(y+z\right)}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}\left(x+y\right)\left(x+z\right)\le\left(\frac{2x+y+z}{2}\right)^2=\frac{\left(2x+y+z\right)^2}{4}\\\left(x+y\right)\left(y+z\right)\le\left(\frac{x+2y+z}{2}\right)^2=\frac{\left(x+2y+z\right)^2}{4}\\\left(x+z\right)\left(y+z\right)\le\left(\frac{x+y+2z}{2}\right)^2=\frac{\left(x+y+2z\right)^2}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{\left(y+z\right)\left(2x+y+z\right)}{\left(x+y\right)\left(x+z\right)}\ge\frac{4\left(y+z\right)\left(2x+y+z\right)}{\left(2x+y+z\right)^2}=\frac{4\left(y+z\right)}{2x+y+z}\\\frac{\left(x+z\right)\left(x+2y+z\right)}{\left(x+y\right)\left(y+z\right)}\ge\frac{4\left(x+z\right)\left(x+2y+z\right)}{\left(x+2y+z\right)^2}=\frac{4\left(x+z\right)}{x+2y+z}\\\frac{\left(x+y\right)\left(x+y+2z\right)}{\left(x+z\right)\left(y+z\right)}\ge\frac{4\left(x+y\right)\left(x+y+2z\right)}{\left(x+y+2z\right)^2}=\frac{4\left(x+y\right)}{x+y+2z}\end{matrix}\right.\)
\(\Rightarrow VT\ge\frac{4\left(y+z\right)}{2x+y+z}+\frac{4\left(x+z\right)}{x+2y+z}+\frac{4\left(x+y\right)}{x+y+2z}\)
\(\Rightarrow VT\ge4\left(\frac{y+z}{2x+y+z}+\frac{x+z}{x+2y+z}+\frac{x+y}{x+y+2z}\right)\)
Ta có: \(x+y+z=1\)
\(\Rightarrow\left\{\begin{matrix}y+z=1-x\\x+z=1-y\\x+y=1-z\end{matrix}\right.\) ( 1 )
\(\Rightarrow\left\{\begin{matrix}2x+y+z=1+x\\x+2y+z=1+y\\x+y+2z=1+z\end{matrix}\right.\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\ge4\left(\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}\right)\)
\(\Rightarrow VT\ge4\left(\frac{1+x-2x}{1+x}+\frac{1+y-2y}{1+y}+\frac{1+z-2z}{1+z}\right)\)
\(\Rightarrow VT\ge4\left[3-\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\right]\)
\(\Rightarrow VT\ge12-4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\)
Chứng minh rằng \(12-4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\ge6\)
\(\Leftrightarrow4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\le6\)
\(\Leftrightarrow\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\le\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}\le\frac{3}{4}\)
\(\Leftrightarrow\frac{1+x-1}{1+x}+\frac{1+y-1}{1+y}+\frac{1+z-1}{1+z}\le\frac{3}{4}\)
\(\Leftrightarrow1-\frac{1}{1+x}+1-\frac{1}{1+y}+1-\frac{1}{1+z}\le\frac{3}{4}\)
\(\Leftrightarrow3-\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\right)\le\frac{3}{4}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge\frac{\left(1+1+1\right)^2}{3+x+y+z}=\frac{9}{4}\)
\(\Rightarrow3-\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\right)\le3-\frac{9}{4}\)
\(\Rightarrow3-\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\right)\le\frac{3}{4}\) ( đpcm )
Vì \(12-4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\ge6\)
\(\Rightarrow VT\ge6\)
\(\Leftrightarrow\)\(\frac{1-x^2}{x+yz}+\frac{1-y^2}{y+xz}+\frac{1-z^2}{z+xy}\ge6\) ( đpcm )
Cách khác:
\(A=\frac{1-x^2}{x+yz}+\frac{1-y^2}{y+xz}+\frac{1-z^2}{z+xy}=\frac{1-x^2}{x(x+y+z)+yz}+\frac{1-y^2}{y(x+y+z)+xz}+\frac{1-z^2}{z(x+y+z)+xy}\)
\(\Leftrightarrow A=\frac{1-x^2}{(x+y)(x+z)}+\frac{1-y^2}{(y+z)(y+x)}+\frac{1-z^2}{(z+x)(z+y)}=\frac{2(x+y+z)-[xy(x+y)+yz(y+z)+xz(x+z)]}{(x+y)(y+z)(x+z)}\)
Có \(A\geq 6\Leftrightarrow 2-[xy(x+y)+yz(y+z)+xz(x+z)]\ge 6(x+y)(y+z)(x+z)\)
\(\Leftrightarrow 2+9xyz\geq 7(x+y+z)(xy+yz+xz)\)
\(\Leftrightarrow 2+9xyz\geq 7(xy+yz+xz)\) \((\star)\)
Theo BĐT Schur bậc 3 kết hợp AM-GM:
\(xyz\geq (x+y-z)(y+z-x)(x+z-y)=(1-2x)(1-2y)(1-2z)\)
\(\Leftrightarrow 9xyz\geq 4(xy+yz+xz)-1\)
\(\Rightarrow 2+9(xy+yz+xz)\geq 1+4(xy+yz+xz)=(x+y+z)^2+4(xy+yz+xz)\)\(\geq 7(xy+yz+xz)\)
Do đó \((\star)\) được CM. Bài toán hoàn tất. Dấu bằng xảy ra khi \(x=y=z=\frac{1}{3}\)