ta có \(\sqrt{\left(a+c\right)\left(a+b\right)}\ge a+\sqrt{bc}\left(1\right)\)

thật vậy \(\left(1\right)\Leftrightarrow\left(a+c\right)\left(a+b\right)\ge a^2+2a\sqrt{bc}+bc\)

\(\Leftrightarrow ab+ac\ge2a\sqrt{bc}\Leftrightarrow b+c\ge2\sqrt{bc}\)(đúng theo BĐT cosi)

cminh tương tự \(\Rightarrow\sqrt{\left(b+c\right)\left(b+a\right)}\ge b+\sqrt{ac};\sqrt{\left(c+a\right)\left(c+b\right)}\ge c+\sqrt{ab}\)

\(\Rightarrow\dfrac{a}{\sqrt{\left(a+c\right)\left(a+b\right)}}\le\dfrac{a}{a+\sqrt{bc}}=\dfrac{1}{1+\dfrac{\sqrt{bc}}{a}}\)

\(tt\Rightarrow P\le\dfrac{1}{1+\dfrac{\sqrt{bc}}{a}}+\dfrac{1}{1+\dfrac{\sqrt{ac}}{b}}+\dfrac{1}{1+\dfrac{\sqrt{ab}}{c}}\)

\(đặt\left(\dfrac{\sqrt{bc}}{a};\dfrac{\sqrt{ac}}{b};\dfrac{\sqrt{ab}}{c}\right)=\left(x;y;z\right)\Rightarrow xyz=1\)

\(\Rightarrow P\le\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\)

ta đi chứng minh \(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\le\dfrac{3}{2}\)

\(\Leftrightarrow2\left(y+1\right)\left(z+1\right)+2\left(x+1\right)\left(z+1\right)+2\left(x+1\right)\left(y+1\right)\le3\left(x+1\right)\left(y+1\right)\left(z+1\right)\)

\(\Leftrightarrow2xy+2xz+2yz+4x+4y+4z+6\le3xyz+3+3xy+3xz+3yz+3x+3y+3z\)

ủa đến đây theo cách làm bth đúng rồi mà sao không ra nhỉ bạn xem lại hộ mình giống bài  n ày mình từng làm r

https://hoc24.vn/vip/289470733648/page-12

\(N=x^2-2xy+y^2+y^2-x=\left(x-y\right)^2-2.\dfrac{1}{2}\left(x-y\right)+\dfrac{1}{4}+y^2-y-\dfrac{1}{4}=\left(x-y-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2-\dfrac{1}{2}\ge-\dfrac{1}{2}\)

\(dấu"="\Leftrightarrow\left\{{}\begin{matrix}x-y-\dfrac{1}{2}=0\\y=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow x=1'y=\dfrac{1}{2}\)

\(A=x^2-2xy+y^2+2y^2-2x+1997=\left(x-y\right)^2-2\left(x-y\right)+1+2y^2-2y+1+1995=\left(x-y-1\right)^2+2\left(y-\dfrac{1}{2}\right)^2+1995+\dfrac{1}{2}\ge1995;5\)

\(dấu"="\Leftrightarrow\left\{{}\begin{matrix}x-y-1=0\\y=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{3}{2};y=\dfrac{1}{2}\)

\(\left(\sqrt{a}+\sqrt{b}\right)^4\le\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}-\sqrt{b}\right)^4=2a^2+2b^2+6ab\)

\(tt\Rightarrow B\le6\left(a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd\right)=6\left(a+b+c+d\right)^2\le6\)

\(dấu"='\Leftrightarrow a=b=c=d=\dfrac{1}{4}\)

\(2.\) \(x^2+3y^2+2xy-10x-14y+10=0\\\)

\(\Leftrightarrow x^2+2xy+y^2+2y^2-10x-14y+10=0\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right).5+25+2y^2-4y+2=17\)

\(\Leftrightarrow\left(x+y-5\right)^2+2\left(y-1\right)^2=17\)

\(\Leftrightarrow-\sqrt{17}\le x+y-5\le\sqrt{17}\Leftrightarrow5-\sqrt{17}\le x+y\le5+\sqrt{15}\)

\(3;\) \(\dfrac{3x^2}{2}+y^2+z^2+yz=1\)

\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\Leftrightarrow\left(x+y+z\right)^2+\left(x-z\right)^2+\left(x-y^2\right)=2\Leftrightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)

\(1;\) \(P=x+y+1\Rightarrow x=P-y-1\)

\(\Rightarrow\left(P-y-1\right)^2+3y^2+2y\left(P-y-1\right)+7\left(P-y-1+y\right)+2y^2+10=0\)

\(\Leftrightarrow\left(P-y-1\right)^2+3y^2+2yP-2y+7P-7+10=0\)

\(\Leftrightarrow P^2+4y^2+2y\left(P-1\right)+7P+4=0\)

\(\Delta'=\left(P-1\right)^2-4\left(P^2+7P+4\right)\ge0\)

\(\Leftrightarrow-3P^2-30P-15\ge0\Leftrightarrow-5-2\sqrt{5}\le P\le5+2\sqrt{5}\)

\(\Rightarrow-5-2\sqrt{5}\le x+y+1\le5+2\sqrt{5}\)

2

\(M=\left(x-2y+1\right)^2+\left(2x+ay+5\right)^2\ge0\)

\(dấu"="\Leftrightarrow\left\{{}\begin{matrix}x-2y+1=0\\2x+ay+5=0\end{matrix}\right.\) \(có\) \(nghiệm\)\(\Leftrightarrow2\ne\dfrac{a}{-2}\Leftrightarrow a\ne4\Rightarrow min=0\)

\(a=-4\Rightarrow M=\left(x-2y+1\right)^2+\left(2x-4y+5\right)^2\)

\(=\left(x-2y+1\right)^2+4\left(x-2y+\dfrac{5}{2}\right)^2\)

\(đặt:x-2y+1=t\Leftrightarrow M=t^2+4\left(t+\dfrac{3}{2}\right)^2=5t^2+12t+9=5\left(t+\dfrac{6}{5}\right)^2+\dfrac{9}{5}\ge\dfrac{9}{5}\)

\(dấu"="\Leftrightarrow t=-\dfrac{6}{5}=x-2y+1\)

\(\Leftrightarrow x-2y=-\dfrac{11}{5}\Rightarrow\left(x;y\right)=\left(b;\dfrac{b+\dfrac{11}{5}}{2}\right)\)\(\Rightarrow min=\dfrac{9}{5}\)

cách giải sai 

\(a+\dfrac{1}{a}=\dfrac{1}{a}+\dfrac{a}{4}+\dfrac{3a}{4}\ge2\sqrt{\dfrac{1}{a}.\dfrac{a}{4}}+\dfrac{3.2}{4}=\dfrac{5}{2}\)

\(dấu""="\Leftrightarrow a=2\)

\(x^2=mx+1\Leftrightarrow x^2-mx-1=0\)

\(\Rightarrow\Delta=m^2+4>0\left(\forall m\right)\)> (P) cắt (d) tại 2 điểm p/b (do ptr có 2 nghiệm p/b A và B

\(A\left(x1;x1^2\right);B\left(x2;x2^2\right)\)

\(\Rightarrow OA=\sqrt{x1^2+x1^4};OB=\sqrt{x2^2+x2^4}\)

\(AB=\sqrt{\left(x1-x2\right)^2+\left(x1^2-x2^2\right)^2}\)

\(\Rightarrow AB^2=x1^2-2x1x2+x2^2+x1^4-2x1^2x2^2+x2^4=x1^2+x2^2+x1^4+x2^4-2\left(-1\right)+2.1=x^2+x2^2+x1^4+x2^4=OA^2+OB^2\)

\(\Rightarrow\Delta OAB\) vuông tại O

\(\Rightarrow Soab=\dfrac{1}{2}OA.OB=\dfrac{1}{2}\sqrt{\left(x1^2+x1^4\right)\left(x2^2+x2^4\right)}=\dfrac{1}{2}\sqrt{\left(x1x2\right)^2+\left(x1.x2\right)^2x2^2+\left(x1x2\right)^2x1^2+\left(x1x2\right)^4}=\dfrac{1}{2}\sqrt{\left(x1+x2\right)^2-2x1x2+2}=\dfrac{1}{2}\sqrt{m^2+4}\)

\(AB\ge\sqrt{m^2+4}\)

\(\Leftrightarrow x1^2+x2^2+x1^4+x1^4\ge m^2+4\)

\(\Leftrightarrow\)\(\left(x1+x2\right)^2-2x1x2+\left(x1^2+x2^2\right)^2-2\left(x1x2\right)^2\ge m^2+4\)

\(\Leftrightarrow m^2+\left[\left(x1+x2\right)^2-2x1x2\right]^2\ge m^2+4\)

\(\Leftrightarrow\left[m^2+2\right]^2\ge4\)

\(\Leftrightarrow m^4+2m^2+4\ge4\Leftrightarrow m^4+m^2\ge0\left(đúng\right)\)

\(\Rightarrowđpcm\)

\(P=\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-\dfrac{5}{ab+bc+ca}=\dfrac{a^2}{ab}+\dfrac{b^2}{bc}+\dfrac{c^2}{ca}-\dfrac{5}{ab+bc+ca}\ge\dfrac{\left(a+b+c\right)^2-5}{ab+bc+ca}=\dfrac{4}{ab+bc+ca}\ge\dfrac{4.3}{\left(a+b+c\right)^2}=\dfrac{12}{3^2}=\dfrac{4}{3}\left(đpcm\right)\)

\(dấu"="\Leftrightarrow a=b=c=1\)

\(x^2=\left(m+1\right)x-m\Leftrightarrow x^2-\left(m+1\right)x+m=0\left(1\right)\)

\(\Rightarrow\Delta>0\Leftrightarrow\left(m+1\right)^2-4m>0\Leftrightarrow m^2-2m+1>0\Leftrightarrow\left(m-1\right)^2>0\Leftrightarrow m\ne1\)

\(\left\{{}\begin{matrix}x1+x2=m+1\\x1.x2=m\end{matrix}\right.\)\(\Rightarrow A\left(x1;y1\right)=\left(x1;\left(m+1\right)x1-m\right)\)

\(\Rightarrow B\left(x2;y2\right)=\left(x2;\left(m+1\right)x2-m\right)\)

\(y1+y2=\left(m+1\right)x1-m+\left(m+1\right)x2-m=\left(m+1\right)\left(x1+x2\right)-2m=\left(m+1\right)^2-2m=m^2+1\ge1\)

\(\Rightarrow min=1\Leftrightarrow m=0\)