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Ta thấy: \(\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}=\Sigma_{cyc}\frac{a^2+bc}{\sqrt[3]{\left(a^2b+b^2c\right)\left(bc^2+ca^2\right)\left(c^2a+ab^2\right)}}\)
Ta lại có: \(\sqrt[3]{\left(a^2b+b^2c\right)\left(bc^2+ca^2\right)\left(c^2a+ab^2\right)}\le\frac{\left(a^2b+b^2c\right)+\left(bc^2+ca^2\right)+\left(c^2a+ab^2\right)}{3}=\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)
\(\Leftrightarrow\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}\ge\frac{\Sigma_{cyc}\left(a^2+bc\right)}{\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)}=\frac{a^2+b^2+c^2+ab+bc+ca}{\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)}\)
Nhận thấy: \(A=\left(a+b+c\right)\left(a^2+b^2+c^2+ab+bc+ca\right)=a^3+b^3+c^3+3abc+2\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)
Theo Schur: \(a^3+b^3+c^3+3abc\ge\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)
\(\Leftrightarrow A\ge3\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}\ge\frac{3\Sigma_{cyc}\left(ab\left(a+b\right)\right)}{\frac{1}{3}\left(a+b+c\right)\Sigma_{cyc}\left(ab\left(a+b\right)\right)}=\frac{9}{a+b+c}\)
Ta có:
\(a+b+\sqrt{2\left(a+c\right)}=a+b+\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a+c}{2}}\ge3\sqrt[3]{\frac{\left(a+b\right)\left(a+c\right)}{2}}\)
Hoàn toàn tương tự ta có:
\(\frac{1}{\left(b+c+\sqrt{2\left(b+a\right)}\right)^3}\le\frac{2}{27\left(b+c\right)\left(b+a\right)}\);
\(\frac{1}{\left(c+b+\sqrt{\left(c+b\right)}\right)^3}\le\frac{2}{27\left(c+a\right)\left(c+b\right)}\)
Cộng theo bất đẳng thức trên ta được:
\(\frac{1}{\left(a+b+\sqrt{2\left(a+c\right)}\right)^3}+\frac{1}{\left(b+c+\sqrt{2\left(b+a\right)}\right)^3}+\frac{1}{\left(c+a+\sqrt{2\left(c+b\right)}\right)^3}\)
\(\le\frac{4\left(a+b+c\right)}{27\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Do đó:
\(\frac{1}{\left(a+b+\sqrt{2\left(a+c\right)}\right)^3}+\frac{1}{\left(b+c+\sqrt{2\left(b+a\right)}\right)^3}+\frac{1}{\left(c+a+\sqrt{2\left(c+b\right)}\right)^3}\)
\(\le\frac{1}{6\left(ab+bc+ca\right)}\)
Vậy bất đẳng thức được chứng minh, bất đẳng thức xày ra khi \(a=b=c=\frac{1}{4}\)
Từ \(a^2+b^2+c^2=3\Rightarrow a+b+c\le3\)
Ta có: \(\sqrt{\dfrac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\dfrac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\dfrac{9}{\left(c+a\right)^2}+b^2}\)
\(\ge\sqrt{9\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)^2+\left(a+b+c\right)^2}\)
\(\ge\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\)
Cần chứng minh \(\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\ge\dfrac{3\sqrt{13}}{2}\)
\(\Leftrightarrow9\left(\dfrac{9}{2t}\right)^2+t^2\ge\dfrac{117}{4}\left(t=a+b+c\le3\right)\)
\(\Leftrightarrow\dfrac{\left(t-3\right)\left(2t-9\right)\left(t+3\right)\left(2t+9\right)}{4t^2}\ge0\)*Đúng*
B1:a)ĐK: \(x\ne 0;4;9\)
b)\(P=\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1+1}{\sqrt{x}+1}\right)\)
\(=\dfrac{x-9-x+4+x^{\dfrac{1}{2}}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{x^{\dfrac{1}{2}}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+1}{x^{\dfrac{1}{2}}}\)
\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1}{x^{\dfrac{1}{2}}}\)\(=\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}\)
c)Vì \(x^{\dfrac{1}{2}}+1>0\forall x\) nên
\(P< 0< =>x-2x^{\dfrac{1}{2}}< 0\)
\(\Leftrightarrow x^{\dfrac{1}{2}}\left(x^{\dfrac{1}{2}}-2\right)< 0\)
\(\Leftrightarrow0< x< 4\)
Vậy 0<x<4 thì P<0
d)tA CÓ: \(\dfrac{1}{P}=\dfrac{x-2x^{\dfrac{1}{2}}}{x^{\dfrac{1}{2}}+1}=\dfrac{x-2x^{\dfrac{1}{2}}+1-1}{x^{\dfrac{1}{2}}+1}=\dfrac{\left(x^{\dfrac{1}{2}}-1\right)^2-1}{x^{\dfrac{1}{2}}+1}\ge-1\)
"=" khi x=1
B2:
a)\(A=x^2-2xy+y^2+4x-4y-5\)
\(=\left(x-y\right)^2+4\left(x-y\right)-5\)
\(=\left(x-y\right)^2-1+4\left(x-y\right)-4\)
\(=\left(x-y+1\right)\left(x-y-1\right)+4\left(x-y-1\right)\)
\(=\left(x-y+5\right)\left(x-y-1\right)\)
b)\(P=x^4+2x^3+3x^2+2x+1\)
\(=\left(x^4+2x^3+x^2\right)+2\left(x^2+x\right)+1\)
\(=\left(x^2+x\right)^2+2\left(x^2+x\right)+1\)
\(=\left(x^2+x+1\right)^2\ge0\forall x\)
Vậy MinP=0
c)\(Q=x^6+2x^5+2x^4+2x^3+2x^2+2x+1\)
\(=\left(x^2+x-1\right)\left(x^4+x^3+2x^2+x+3\right)+4\)
\(=\left(1-1\right)\left(x^4+x^3+2x^2+x+3\right)+4\)
\(=0\left(x^4+x^3+2x^2+x+3\right)+4=4\)
Vậy x^2+x=1 thì Q=4
B3:a)\(2xy+x+y=83\)
\(\Leftrightarrow x\left(2y+1\right)+\dfrac{1}{2}\left(2y+1\right)=\dfrac{167}{2}\)
\(\Leftrightarrow2x\left(2y+1\right)+1\left(2y+1\right)=167\)
\(\Leftrightarrow\left(2x+1\right)\left(2y+1\right)=167\)
Mà \(Ư\left(167\right)=\left\{\pm1;\pm167\right\}\)
\(\Leftrightarrow\left(x;y\right)=\left(-84;-1\right);\left(-1;-84\right);\left(0;83\right);\left(83;0\right)\)
Vậy...
b)\(y^2+2xy-3x-2=0\)
\(\Leftrightarrow x^2+y^2+2xy-x^2-3x-2=0\)
\(\Leftrightarrow\left(x+y\right)^2=x^2+3x+2\)
\(\Leftrightarrow\left(x+y\right)^2=\left(x+1\right)\left(x+2\right)\)
Vì \(x;y\in Z\) nên VT là số chính phương VP là tích 2 số nguyên liên tiếp
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2\end{matrix}\right.\)
Vậy...
B5:\(B=\dfrac{x^2+x+1}{x^2-x+1}\)
\(\Leftrightarrow x^2\left(B-1\right)+x\left(-B-1\right)+\left(B-1\right)=0\)
\(\Delta=\left(-B-1\right)^2-4\left(B-1\right)\left(B-1\right)\)
\(=-\left(B-3\right)\left(3B-1\right)\)
pt có nghiệm khi \(\Delta\ge0\)
\(\Leftrightarrow\left(B-3\right)\left(3B-1\right)\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}B-3\le0\\3B-1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}B\le3\\B\ge\dfrac{1}{3}\end{matrix}\right.\)
Min B=1/3 khi x=-1; Max B=3 khi x=1
Đề bài hình như bị sai em, thay điểm rơi ko thỏa mãn
Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng
Ta có:\(\left(a^2+bc\right)\left(b+c\right)=b\left(a^2+c^2\right)+c\left(a^2+b^2\right)\)
\(\Rightarrow\sqrt{\frac{\left(a^2+bc\right)\left(b+c\right)}{a\left(b^2+c^2\right)}}=\sqrt{\frac{b\left(a^2+c^2\right)+c\left(a^2+b^2\right)}{a\left(b^2+c^2\right)}}\)
Tương tự\(\Rightarrow\)VT=\(\Sigma\sqrt{\frac{b\left(a^2+c^2\right)+c\left(a^2+b^2\right)}{a\left(b^2+c^2\right)}}\)
Đặt \(x=a\left(b^2+c^2\right)\);\(y=b\left(a^2+c^2\right)\);\(z=c\left(b^2+a^2\right)\)
VT=\(\sqrt{\frac{x+y}{z}}+\sqrt{\frac{y+z}{x}}+\sqrt{\frac{x+z}{y}}\ge3\sqrt[6]{\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}}\ge3\sqrt{2}\)(BĐT Cô-si)
Dấu''='' xra\(\Leftrightarrow\)a=b=c
HD giải, Bài này còn một cách khác là xài bunhia ._. cơ mà rối lắm. :)))
