Bài 1: Căn bậc hai

CM cái j v đề thiếu rồi

Aps dụng bất đẳng thức cô si cho 2 số 1-x và 1-x ta có:

\(\dfrac{1-x+1-z}{2}\ge\sqrt{\left(1-x\right)\left(1-z\right)}\)

\(\Leftrightarrow\left(1-z\right)\left(1-x\right)\le\left(\dfrac{1-z+1-x}{2}\right)^2\)

\(\Leftrightarrow4\left(1-z\right)\left(1-x\right)\le\left(1+y\right)^2\)

\(\Leftrightarrow4\left(1-x\right)\left(1-y\right)\left(1-z\right)\le\left(1+y\right)^2\left(1-y\right)\)

Ta có: \(1-y^2\le1\)

\(\left(1+y\right)^2\left(1-y\right)=\left(1+y\right)\left(1-y\right)^2=\left(x+2y+z\right)\left(1-y\right)^2\)

Do đó: \(4\left(1-x\right)\left(1-y\right)\left(1-z\right)\le x+2y+z\)

Áp dụng BĐT cô-si cho 2 số 1-x và 1-z ta được:

\(\dfrac{1-x+1-z}{2}\ge\sqrt{\left(1-x\right)\left(1-z\right)}\)

\(\Leftrightarrow\text{ ( 1 − x ) ( 1 − z )\le(\dfrac{\text{1 − x + 1 −}z}{2})^2 }\)

\(\Leftrightarrow\text{4 ( 1 − x ) ( 1 − z ) ≤ ( 1 + y ) ^2}\)

\(\Leftrightarrow\text{ 4 ( 1 − x ) ( 1 − z ) ( 1 − y ) ≤ ( 1 + y ) ^2 ( 1 − y )}\)

mặt khác\(\text{ 1 − y ^2 ≤ 1}\)

\(\text{( 1 + y ) ^2 ( 1 − y ) = ( 1 + y ) ( 1 − y ^2) = ( x + 2y + z ) ( 1 − y^2 ) (1+y)^2(1−y)=(1+y)(1−y^2)=(x+2y+z)(1−y^2)}\)Do đó: 4(1−x)(1−y)(1−z)≤x+2y+z

\(\sqrt{x+2\sqrt{x-1}}\)(ĐK:x\(\ge1\))

\(=\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}\)

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

\(=\left|\sqrt{x-1}+1\right|\)

\(=\sqrt{x-1}+1\)

\(\sqrt{x+2\sqrt{x-1}}\)

\(=\sqrt{x-1+2\sqrt{x-1}+1}\)

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

\(=\left|\sqrt{x-1}+1\right|\)

\(=\sqrt{x-1}+1\)

\(\)

ĐKXĐ \(x\ge 1\)

\(\sqrt{x+2\sqrt{x-1} } \)\(=\sqrt{x-1+2\sqrt{x-1}+1 } =\sqrt{(\sqrt{x-1}+1)^2 } =|\sqrt{x-1}+1| =\sqrt{x-1}+1 \)

Chứng minh đẳng thức phụ:

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)

\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{a+b+c}{abc}\)

\(\Rightarrow\) Với \(a+b+c=0\). Ta có: \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)

\(\Leftrightarrow\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\)với \(a+b+c=0\)

Ta có:

\(S=\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+.....+\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}\)

Áp dụng đẳng thức phụ trên:

\(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{1^2}+\dfrac{1}{\left(-2\right)^2}}=1+1-\dfrac{1}{2}\left(>0\right)\)

\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{\left(-3\right)^2}}=1+\dfrac{1}{2}-\dfrac{1}{3}\left(>0\right)\)

\(\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{\left(-4\right)^2}}=1+\dfrac{1}{3}-\dfrac{1}{4}\left(>0\right)\)
\(.................\)

\(\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{99^2}+\dfrac{1}{\left(-100\right)^2}}=1+\dfrac{1}{99}-\dfrac{1}{100}\)

Cộng vế với vế các đẳng thức trên, ta có:

\(S=\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+........+\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}\)

\(=1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+............+1+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=99+1-\dfrac{1}{100}=99+\dfrac{99}{100}=99\dfrac{99}{100}\)

Chứng minh kiểu khác :v

\(\forall n\in\)N^*, ta có:

\(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}\)

\(=\sqrt{\dfrac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\)

\(=\sqrt{\dfrac{\left[n.\left(n+1\right)\right]^2+2n\left(n+1\right)+1}{\left[n\left(n+1\right)\right]^2}}\)

\(=\sqrt{\dfrac{\left[n\left(n+1\right)+1\right]^2}{\left[n\left(n+1\right)\right]^2}}=\dfrac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\dfrac{1}{n\left(n+1\right)}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\)

Việc còn lại là áp dụng vào bài thoy :v

không biết

\(M=\sqrt{\dfrac{a}{b+c+2a}}+\sqrt{\dfrac{b}{c+a+2b}}+\sqrt{\dfrac{c}{a+b+2c}}\)

\(\le\dfrac{1}{4}+\dfrac{a}{b+c+2a}+\dfrac{1}{4}+\dfrac{b}{c+a+2b}+\dfrac{1}{4}+\dfrac{c}{a+b+2c}\)

\(\le\dfrac{3}{4}+\dfrac{1}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)

\(=\dfrac{3}{4}+\dfrac{1}{4}.\left(1+1+1\right)=\dfrac{3}{2}\)

<=> √12.25

<=>a+1+b+1+c+1 12.25

<=>4<12.25(dpcm)

hay

Áp dụng BĐT Bunyakovsky, ta có:

\(\left(a+1+b+1+c+1\right)3\ge\left(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\right)^2\)

\(\Rightarrow\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le\sqrt{12}< 3,5\)