Áp dụng BĐT cauchy-schwarz:

\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)

BĐT cần chứng minh tương đương :

\(\left(a+b+c\right)^2\ge3\left(\sqrt{a^3c}+\sqrt{b^3a}+\sqrt{c^3b}\right)\)

Thật vậy, Áp dụng BĐT \(\left(X+Y+Z\right)^2\ge3\left(XY+YZ+ZX\right)\)

Với \(\left\{{}\begin{matrix}X=a+\sqrt{bc}-\sqrt{ac}\\Y=b+\sqrt{ac}-\sqrt{ab}\\Z=c+\sqrt{ab}-\sqrt{bc}\end{matrix}\right.\) ta có ngay ĐPCM. ( mất chút time khai triển)

Dấu = xảy ra khi X=Y=Z hay a=b=c

Bài này có xuất hiện rồi ,you vào mục tìm kiếm là thấy liền.

Lời giải vắn tắt:

\(A=\sum\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sum\dfrac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(1+ab-c^2\right)}}\ge\sum\dfrac{2\left(ab+2c^2\right)}{1+2ab+c^2}=\sum\dfrac{2\left(ab+2c^2\right)}{\left(a+b\right)^2+2c^2}\ge\sum\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}=\sum\left(ab+2c^2\right)=ab+bc+ca+2\)

( thay \(a^2+b^2+c^2=1\))

Áp dụng BĐT AM-GM:

\(VT=\sum\dfrac{\sqrt{\left(x+y\right)^2-xy}}{4yz+1}\ge\sum\dfrac{\sqrt{\left(x+y\right)^2-\dfrac{1}{4}\left(x+y\right)^2}}{\left(y+z\right)^2+1}=\sum\dfrac{\dfrac{\sqrt{3}}{2}\left(x+y\right)}{\left(y+z\right)^2+1}\)

Set \(\left\{{}\begin{matrix}x+y=a\\y+z=b\\z+x=c\end{matrix}\right.\)thì giả thiết trở thành \(a+b+c=3\) và cần chứng minh \(\dfrac{\sqrt{3}}{2}.\sum\dfrac{a}{b^2+1}\ge\dfrac{3\sqrt{3}}{4}\)

\(\Leftrightarrow\sum\dfrac{a}{b^2+1}\ge\dfrac{3}{2}\)( đến đây quen thuộc rồi)

Ta có:\(\sum\dfrac{a}{b^2+1}=\sum a-\sum\dfrac{ab^2}{b^2+1}\ge3-\sum\dfrac{ab^2}{2b}\)(AM-GM)

\(VT\ge3-\sum\dfrac{ab}{2}\ge3-\dfrac{\dfrac{1}{3}\left(a+b+c\right)^2}{2}=\dfrac{3}{2}\)( AM-GM)

Vậy ta có đpcm.Dấu = xảy ra khi a=b=c=1 hay \(x=y=z=\dfrac{1}{2}\)

\(A=\sum\sqrt{\dfrac{1}{1+a^2}}=\sum\sqrt{\dfrac{bc}{bc+a.abc}}=\sum\sqrt{\dfrac{bc}{bc+a\left(a+b+c\right)}}=\sum\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\sum\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)=\dfrac{3}{2}\)

Áp dụng cauchy-schwarz:

\(\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+e}+\dfrac{d}{e+a}+\dfrac{e}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+bd}+\dfrac{c^2}{cd+ce}+\dfrac{d^2}{ed+ad}+\dfrac{e^2}{ae+be}\ge\dfrac{\left(a+b+c+d\right)^2}{ab+ac+ad+ae+bc+bd+be+cd+ce+de}\)

Giờ chỉ cần chứng minh

\(ab+ac+ad+ae+bc+bd+be+cd+ce+de\le\dfrac{2}{5}\left(a+b+c+d+e\right)^2\)

\(\Leftrightarrow ab+ac+ad+ae+bc+bd+be+cd+ce+de\le2\left(a^2+b^2+c^2+d^2+e^2\right)\)

điều này hiển nhiên đúng theo AM-GM:

\(ab\le\dfrac{a^2+b^2}{2};ac\le\dfrac{a^2+c^2}{2};ad\le\dfrac{a^2+d^2}{2}...\)

Cứ vậy ta thu được đpcm .Dấu = xảy ra khi a=b=c=d=e

P/s: : ]

\(Pt\Leftrightarrow x-\sqrt{\dfrac{x-1}{x}}=\sqrt{\dfrac{x^2-1}{x}}\)( DKXĐ : \(x\ge1\))

\(\Leftrightarrow x=\sqrt{\dfrac{x-1}{x}}\left(\sqrt{x+1}+1\right)=\sqrt{\dfrac{x-1}{x}}.\dfrac{x+1-1}{\sqrt{x+1}-1}\)

\(\Leftrightarrow x=\dfrac{\sqrt{x}.\sqrt{x-1}}{\sqrt{x+1}-1}\)

\(\Leftrightarrow\sqrt{x}.\left(\sqrt{x}-\dfrac{\sqrt{x-1}}{\sqrt{x+1}-1}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loai\right)\\\sqrt{x}=\dfrac{\sqrt{x-1}}{\sqrt{x+1}-1}\end{matrix}\right.\)

Có:\(x=\dfrac{x-1}{x+2-2\sqrt{x+1}}\) (ĐK: \(\sqrt{x+1}>1\))

\(\Leftrightarrow x^2+\left(x+1\right)-2x\sqrt{x+1}=0\)

\(\Leftrightarrow\left(x-\sqrt{x+1}\right)^2=0\)

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

\(\Leftrightarrow x^2-x-1=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\left(chọn\right)\\x=\dfrac{1-\sqrt{5}}{2}\left(loại\right)\end{matrix}\right.\)

Vậy ...

Ta có:

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

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

Áp dụng AM-GM:

\(\dfrac{2ab}{\left(a-b\right)^2}+\dfrac{\left(a-b\right)^2}{ab}\ge2\sqrt{2}\)

Do đó \(VT\ge3+2\sqrt{2}\)

Dấu = xảy ra khi \(\left(a-b\right)^2=2a^2b^2\)

P/s: ăn may

Từ Giả thiết :

\(xy\left(x-y\right)^2=\left(x+y\right)^2\Leftrightarrow xy\left[\left(x+y\right)^2-4xy\right]=\left(x+y\right)^2\)

Set \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\)

\(GT\Leftrightarrow b\left(a^2-4b\right)=a^2\Leftrightarrow a^2=\dfrac{4b^2}{b-1}\)

Dễ dàng nhận thấy

\(\dfrac{b^2}{b-1}+4\left(b-1\right)\ge4b\)

\(\Leftrightarrow\dfrac{b^2}{b-1}\ge4\) ( Dấu = xảy ra khi b=2)

\(\Rightarrow a^2\ge16\Leftrightarrow a\ge4\) ( vì a dương )

hay \(x+y\ge4\)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}x+y=4\\xy=2\end{matrix}\right.\)

Áp dụng BĐT cauchy-schwarz:

\(\sum\dfrac{a^4b}{2a+b}=\sum\dfrac{a^4b^2}{2ab+b^2}\ge\dfrac{\left(a^2b+b^2c+c^2a\right)^2}{\left(a+b+c\right)^2}\)

giờ ta chỉ cần có:\(a^2b+b^2c+c^2a\ge a+b+c\)

Áp dụng AM-GM:

\(a^2b+\dfrac{1}{b}\ge2a\)..tương tự ,ta suy ra:

\(a^2b+b^2c+c^2a\ge2\left(a+b+c\right)-\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)(*)

Theo giả thiết: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le3\)

Dễ dàng suy ra được \(a+b+c\ge3\) ( từ BĐT \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\))

theo đó thì \(a+b+c\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

Nên từ (*) ta có đpcm.

Dấu = xảy ra khi a=b=c=1