§1. Bất đẳng thức

Đặt \(\left\{{}\begin{matrix}x+\sqrt{x^2+1}=a>0\\y+\sqrt{y^2+1}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}=a-x\\\sqrt{y^2+1}=b-y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2ax=a^2-1\\2by=b^2-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a^2-1}{2a}\\y=\dfrac{b^2-1}{2b}\end{matrix}\right.\)

 \(\Rightarrow\left(\dfrac{a^2-1}{2a}+\sqrt{\left(\dfrac{b^2-1}{2b}\right)+1}\right)\left(\dfrac{b^2-1}{2b}+\sqrt{\left(\dfrac{a^2-1}{2a}\right)+1}\right)=1\)

\(\Rightarrow\left(\dfrac{a^2-1}{2a}+\dfrac{b^2+1}{2b}\right)\left(\dfrac{b^2-1}{2b}+\dfrac{a^2+1}{2a}\right)=1\)

\(\Rightarrow\left(\dfrac{a+b}{2}+\dfrac{a-b}{2ab}\right)\left(\dfrac{a+b}{2}-\dfrac{a-b}{2ab}\right)=\dfrac{4ab}{4ab}=\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4ab}\)

\(\Rightarrow\dfrac{\left(a+b\right)^2}{4}-\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4\left(ab\right)^2}+\dfrac{\left(a-b\right)^2}{4ab}=0\)

\(\Rightarrow\dfrac{\left(a+b\right)^2}{4}\left(1-\dfrac{1}{ab}\right)+\dfrac{\left(a-b\right)^2}{4ab}\left(1-\dfrac{1}{ab}\right)=0\)

\(\Rightarrow\left(1-\dfrac{1}{ab}\right)\left(\dfrac{\left(a+b\right)^2}{4}+\dfrac{\left(a-b\right)^2}{4ab}\right)=0\)

\(\Rightarrow1-\dfrac{1}{ab}=0\Rightarrow ab=1\)

\(\Rightarrow\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)

\(\Rightarrow x+y=0\Rightarrow y=-x\)

\(P=2\left(x^2+\left(-x\right)^2\right)+0=4x^2\ge0\)

Dấu "=" xảy ra khi \(x=y=0\)

dễ hiểu mà bạn

Gọi 2 số đó là a và b (a\(\ge0,b\ge0\) )

 câu a

Áp dụng  BĐT Bu-nhia -xkop-ki ,ta có

a+b\(\le\sqrt{\left(a^2+b^2\right)\left(1^2+1^2\right)}\)

\(\Leftrightarrow82\le\sqrt{\left(a^2+b^2\right)2}\) \(\Rightarrow\) \(6724\le\left(a^2+b^2\right)2\Leftrightarrow\left(a^2+b^2\right)\ge3362\)

Vậy Min a²+b²=3362\(\Leftrightarrow a=b=41\)

Giúp e với ; plz 

Ta có :  \(cos^2A+cos^2B+cos^2C=1-2.cosA.cosB.cosC\)  

Đặt cos A = a ; cos B = b ; cos C = c  thì : \(a^2+b^2+c^2+2abc=1\)

Dự đoán : a = b = c = 1/2 nên ta đặt 

a = \(\sqrt{\dfrac{xy}{\left(y+z\right)\left(z+x\right)}}\)    ; \(b=\sqrt{\dfrac{yz}{\left(x+z\right)\left(x+y\right)}};c=\sqrt{\dfrac{xz}{\left(y+z\right)\left(x+y\right)}}\)  ( x ; y ; z > 0 ) 

Khi đó : \(\Sigma\sqrt{\dfrac{cosA.cosB}{cosC}}=\Sigma\sqrt{\dfrac{y}{x+z}}\)  

Cần c/m : \(\Sigma\sqrt{\dfrac{y}{x+z}}>2\)   (*) 

BĐT quen thuộc ; AD BĐT AM - GM ta được : \(\sqrt{\dfrac{x+z}{y}}\le\dfrac{1}{2}\left(\dfrac{x+y+z}{y}\right)\Rightarrow\sqrt{\dfrac{y}{x+z}}\ge\dfrac{2y}{x+y+z}\) 

Suy ra : \(\Sigma\sqrt{\dfrac{y}{x+z}}\ge\dfrac{2\left(x+y+z\right)}{x+y+z}=2\) 

" = " ko xảy ra nên hiển nhiên (*) đúng

Hoàn tất c/m 

Có \(\dfrac{1}{4-\sqrt{ab}}\le\dfrac{1}{4-\dfrac{\sqrt{2\left(a^2+b^2\right)}}{2}}=\dfrac{2}{8-\sqrt{2\left(a^2+b^2\right)}}\)

Tương tự: \(\dfrac{1}{4-\sqrt{bc}}\le\dfrac{2}{8-\sqrt{2\left(b^2+c^2\right)}}\),  \(\dfrac{1}{4-\sqrt{ca}}\le\dfrac{2}{8-\sqrt{2\left(a^2+c^2\right)}}\)

Đặt \(\left(a^2+b^2;b^2+c^2;c^2+a^2\right)=\left(x;y;z\right)\)

Khi đó \(\left\{{}\begin{matrix}x+y+z=6\\z,y,z>0\end{matrix}\right.\) (1)

Đặt VT của bđt là A

Có  \(A=\dfrac{1}{4-\sqrt{ab}}+\dfrac{1}{4-\sqrt{bc}}+\dfrac{1}{4-\sqrt{ca}}\le\dfrac{2}{8-\sqrt{2x}}+\dfrac{2}{8-\sqrt{2y}}+\dfrac{2}{8-\sqrt{2z}}\)

Ta cm bđt phụ: \(\dfrac{2}{8-\sqrt{2x}}\le\dfrac{1}{36}\left(x-2\right)+\dfrac{1}{3}\)

Thật vậy bđt trên tương đương \(\dfrac{6}{3\left(8-\sqrt{2x}\right)}-\dfrac{8-\sqrt{2x}}{3\left(8-\sqrt{2x}\right)}-\dfrac{1}{36}\left(x-2\right)\le0\)

\(\Leftrightarrow\dfrac{\sqrt{2}\left(\sqrt{x}-\sqrt{2}\right)}{3\left(8-\sqrt{2x}\right)}-\dfrac{\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)}{36}\le0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{2}\right)\left[\dfrac{\sqrt{2}.12}{36\left(8-\sqrt{2x}\right)}-\dfrac{\left(\sqrt{x}+\sqrt{2}\right)\left(8-\sqrt{2x}\right)}{36\left(8-\sqrt{2x}\right)}\right]\le0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{2}\right)^2.\dfrac{\left(\sqrt{x}-2\sqrt{2}\right)}{36\left(8-\sqrt{2x}\right)}\le0\)  (*)

Từ (1) ta có \(x\in\left(0;6\right)\) nên bđt phụ trên luôn đúng
Tương tự ta cũng có \(\dfrac{2}{8-\sqrt{2y}}\le\dfrac{1}{36}\left(y-2\right)+\dfrac{1}{3}\) , \(\dfrac{2}{8-\sqrt{2z}}\le\dfrac{1}{36}\left(z-2\right)+\dfrac{1}{3}\)
Từ đó => \(A\le\dfrac{1}{36}\left(x+y+z-6\right)+1=\dfrac{1}{36}\left(6-6\right)+1=1\) (đpcm)
Dấu = xảy ra <=> x=y=z=2 <=> a=b=c=1

Sử dụng Dirichlet và biến đổi là được, không cần tới Schur:

Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có ít nhất 2 số cùng phía so với 2, ko mất tính tổng quát, giả sử đó là a và b

\(\Rightarrow\left(a-2\right)\left(b-2\right)\ge0\Rightarrow ab\ge2a+2b-4\Rightarrow2abc\ge4ac+4bc-8c\)

\(\Rightarrow P\ge3\left(a^2+b^2+c^2\right)+4ac+4bc-8c=2\left(a+b+c\right)^2+\left(a-b\right)^2+c^2-8c-2ab\)

\(\Rightarrow P\ge c^2-8c-2ab+72\ge c^2-8c-\dfrac{1}{2}\left(a+b\right)^2+72\)

\(\Rightarrow P\ge c^2-8c-\dfrac{1}{2}\left(6-c\right)^2+72=\dfrac{1}{2}\left(c-2\right)^2+52\ge52\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Áp dụng BĐT BSC:

\(A=\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}\)

\(=\dfrac{\dfrac{1}{16}}{x}+\dfrac{\dfrac{1}{4}}{y}+\dfrac{1}{z}\)

\(\ge\dfrac{\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2}{x+y+z}=\dfrac{49}{16}\)

\(minA=\dfrac{49}{16}\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\dfrac{1}{4}}{x}=\dfrac{\dfrac{1}{2}}{y}=\dfrac{1}{z}\\x+y+z=1\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y;z\right)=\left(\dfrac{1}{7};\dfrac{2}{7};\dfrac{4}{7}\right)\)

\(P=\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}+\dfrac{49}{16}-\dfrac{49}{16}\)

\(P=\left(\dfrac{1}{16x}+\dfrac{49x}{16}\right)+\left(\dfrac{1}{4y}+\dfrac{49y}{16}\right)+\left(\dfrac{1}{z}+\dfrac{49z}{16}\right)-\dfrac{49}{16}\)

\(P\ge2\sqrt{\dfrac{49x}{16x.16}}+2\sqrt{\dfrac{49y}{4y.16}}+2\sqrt{\dfrac{49z}{z.16}}-\dfrac{49}{16}=\dfrac{49}{16}\)

Dấu "=" xảy ra khi...

Lời giải:

a) BĐT cần CM tương đương với:

\(\sqrt{a(a+1)}>1+\sqrt{a(a-1)}\)

\(\Leftrightarrow a(a+1)>1+a(a-1)+2\sqrt{a(a-1)}\)

\(\Leftrightarrow a+(a-1)-2\sqrt{a(a-1)}>0\)

\(\Leftrightarrow (\sqrt{a}-\sqrt{a-1})^2>0\) (luôn đúng với mọi $a\geq 1$)

Do đó ta có đpcm.

b) ĐK cần sửa lại là $x>6$ và dấu "=" không xảy ra bạn nhé

\(\frac{x+8}{\sqrt{x-6}}=\frac{(x-6)+14}{\sqrt{x-6}}\geq \frac{2\sqrt{14(x-6)}}{\sqrt{x-6}}=2\sqrt{14}>6\)

Ta có đpcm.

Ta có: \(\left(x-y\right)\left(1-xy\right)\le\dfrac{1}{4}\left(x-y+1-xy\right)^2=\dfrac{1}{4}\left(x+1\right)^2\left(1-y\right)^2\)

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

\(P_{max}=\dfrac{1}{4}\) khi \(\left(x;y\right)=\left(1;0\right)\)

Lại có:

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

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

\(\Rightarrow-P\le\dfrac{1}{4}\Rightarrow P\ge-\dfrac{1}{4}\)

\(P_{min}=-\dfrac{1}{4}\) khi \(\left(x;y\right)=\left(0;1\right)\)

(Do \(y\ge0\Rightarrow\dfrac{4y}{y^2+2y+1}\ge0\Rightarrow1-\dfrac{4y}{y^2+2y+1}\le1\Rightarrow...\))

1 bài Mincopxki khá quen:

\(P\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{\left(a+b+c\right)^2}}\)

Đến đây thì nó là bài Cô-si có biên, cứ tách ghép theo điểm rơi là được:

\(P\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{16\left(a+b+c\right)^2}+\dfrac{1215}{16\left(a+b+c\right)^2}}\)

\(P\ge\sqrt{2\sqrt{\dfrac{81\left(a+b+c\right)^2}{16\left(a+b+c\right)^2}}+\dfrac{1215}{16.\left(\dfrac{3}{2}\right)^2}}=\dfrac{3\sqrt{17}}{2}\)

Dấu "=" xayr a khi \(a=b=c=\dfrac{1}{2}\)