Ta có đánh giá sau với a không âm:

\(\dfrac{a}{1+a^2}\le\dfrac{36a+3}{50}\)

Thật vậy, BĐT tương đương:

\(\left(36a+3\right)\left(a^2+1\right)\ge50a\)

\(\Leftrightarrow\left(3a-1\right)^2\left(4a+3\right)\ge0\) (luôn đúng)

Tương tự: \(\dfrac{b}{1+b^2}\le\dfrac{36b+3}{50}\) ; \(\dfrac{c}{1+c^2}\le\dfrac{36c+3}{50}\)

Cộng vế: \(VT\le\dfrac{36\left(a+b+c\right)+9}{50}=\dfrac{9}{10}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Ta chứng minh bđt phụ \(\dfrac{a}{1+a^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\)

Thật vậy bđt trên \(\Leftrightarrow\dfrac{-3a^2+10a-3}{10\left(1+a^2\right)}-\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\le0\)

\(\Leftrightarrow\left(a-\dfrac{1}{3}\right)\left[\dfrac{3\left(3-a\right)}{10\left(1+a^2\right)}-\dfrac{18}{25}\right]\le0\)

\(\Leftrightarrow-\dfrac{36\left(a-\dfrac{1}{3}\right)^2\left(\dfrac{3}{4}+a\right)}{50\left(1+a^2\right)}\le0\) ( luôn đúng với mọi \(a\)\(\ge\)0)

Tương tự cũng có:\(\dfrac{b}{1+b^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(b-\dfrac{1}{3}\right)\); \(\dfrac{c}{1+c^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(c-\dfrac{1}{3}\right)\)

Cộng vế với vế => VT\(\le\dfrac{9}{10}+\dfrac{18}{25}\left(a+b+c-1\right)=\dfrac{9}{10}\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Đặt \(T=\left(a+b\right)\left(b+c\right)\left(c+a\right)>0\)

\(BDT\Leftrightarrow\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a+b}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2+bc}{b+c}-a+\dfrac{b^2+ca}{c+a}-b+\dfrac{c^2+ab}{a+b}-c\ge0\)

\(\Leftrightarrow\dfrac{a^2+bc-ab-ac}{b+c}+\dfrac{b^2+ac-ab-bc}{a+c}+\dfrac{c^2+ab-ac-bc}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{b+c}+\dfrac{\left(b-a\right)\left(b-c\right)}{a+c}+\dfrac{\left(c-a\right)\left(c-b\right)}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)\left(a^2-c^2\right)+\left(b^2-a^2\right)\left(b^2-c^2\right)+\left(c^2-a^2\right)\left(c^2-b^2\right)}{T}\ge0\)

\(\Leftrightarrow\dfrac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{T}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2}{2T}\ge0\)

Xảy ra khi \(a=b=c\)

\(BĐT\Leftrightarrow\sum\left(\dfrac{1}{a}-\dfrac{b+c}{a^2+bc}\right)\ge0\)

\(\Leftrightarrow\sum\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\)

Giả sử \(a\ge b\ge c\)thì

\(\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\).vậy nên chỉ cần chứng minh

\(\dfrac{\left(b-c\right)\left(b-a\right)}{b\left(b^2+ac\right)}+\dfrac{\left(c-a\right)\left(c-b\right)}{c\left(c^2+ab\right)}\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\dfrac{b-a}{b\left(b^2+ac\right)}+\dfrac{a-c}{c\left(c^2+ab\right)}\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\left(b-a\right)\left(c^3+abc\right)+\left(a-c\right)\left(b^3+abc\right)\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)^2\left(b+c\right)\left(ab+ac-bc\right)\ge0\)( đúng vì \(a\ge b\ge c\))

Vậy BĐT được chứng minh.

Dấu = xảy ra khi a=b=c

2a)

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2a+b+c}=\dfrac{1}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{1}{a+2b+c}=\dfrac{1}{a+b+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{1}{a+b+2c}=\dfrac{1}{a+c+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)

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

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

\(\Rightarrow VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)

Chứng minh rằng \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)

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

\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )

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

Mà \(VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)

\(\Rightarrow\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)( đpcm )

Dấu " = " xảy ra khi \(a=b=c\)

2b)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}1+a^2\ge2\sqrt{a^2}=2a\\1+b^2\ge2\sqrt{b^2}=2b\\1+c^2\ge2\sqrt{c^2}=2c\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\\\dfrac{b}{1+b^2}\le\dfrac{b}{2b}=\dfrac{1}{2}\\\dfrac{c}{1+c^2}\le\dfrac{c}{2c}=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\le\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=1\)

Bài 1)

Nháp : nhìn nhanh ta thấy nên áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Giải

Vì x,y > 0 =) 2x + y > 0 , x + 2y > 0

Áp dụng BĐT cauchy dạng phân thức cho hai bộ số không âm \(\dfrac{1}{2x+y}\)và\(\dfrac{1}{x+2y}\)

\(\Rightarrow\dfrac{1}{x+2y}+\dfrac{1}{2x+y}\ge\dfrac{4}{x+2y+2x+y}=\dfrac{4}{3\left(x+y\right)}\)

\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3\left(x+y\right)}=4\)

Dấu '' = "xảy ra khi và chỉ khi x + 2y = y + 2x (=) x=y

ta có:

\(1+a^2\ge2\sqrt{a^2.1}=2a\Rightarrow\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\)

tương tự, ta có: \(\dfrac{b}{b^2+1}\le\dfrac{1}{2}\); \(\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)

công từ vế với nhau, ta có điều cần phải chứng minh

a.

Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)

Tương tự: \(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)

Cộng vế:

\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

b.

Ta có:

\(a^2+bc\ge2\sqrt{a^2bc}=2\sqrt{ab.ac}\Rightarrow\dfrac{1}{a^2+bc}\le\dfrac{1}{2\sqrt{ab.ac}}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{ac}\right)\)

Tương tự: \(\dfrac{1}{b^2+ac}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{bc}\right)\) ; \(\dfrac{1}{c^2+ab}\le\dfrac{1}{4}\left(\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)

Cộng vế với vế:

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

Dấu "=" xảy ra khi \(a=b=c\)

\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+ac+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)=\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\) Chứng minh tương tự ta được:

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

\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+a}+\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)=\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\left(1+1+1\right)=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)

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

Tương tự: \(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)

Cộng vế:

\(VT\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

Đặt \(\left\{{}\begin{matrix}x=a+b+c\\y=ab+bc+ca\end{matrix}\right.\) khi đó \(BDT\Leftrightarrow\dfrac{x^2+4x+y+3}{x^2+2x+y+xy}\le\dfrac{12+4x+y}{9+4x+2y}\)

\(\Leftrightarrow\dfrac{x^2+4x+y+3}{x^2+2x+y+xy}-1\le\dfrac{12+4x+y}{9+4x+2y}-1\)

\(\Leftrightarrow\dfrac{2x+3-xy}{x^2+2x+y+xy}\le\dfrac{3-y}{9+4x+2y}\)

\(\Leftrightarrow\dfrac{5x^2-3x^2y-xy^2-6xy+24x+y^2+3y+27}{\left(4x+2y+9\right)\left(x^2+xy+2x+y\right)}\le0\)

Đúng vì \(\dfrac{5}{3}x^2y\ge5x^2;\dfrac{x^2y}{3}\ge y^2;xy^2\ge9x;5xy\ge15x;xy\ge3y;x^2y\ge27\)

1.

Áp dụng BĐT Cauchy-Schwarz:

\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)

Tương tự:

\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)

\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)

Cộng vế:

\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)

Dấu "=" xảy ra khi \(a=b=c\)

2.

Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)

Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)

Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)

Biến đổi giả thiết:

\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)

\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)

\(\Rightarrow ab+bc+ca=a+b+c-1\)

BĐT cần chứng minh trở thành:

\(a^2+b^2+c^2\ge1\)

\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)

\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)

\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)

Xét hiệu VT - VP

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

Do a,b,c bình đẳng nên giả sử a\(\ge\)b\(\ge\)c, khi đó \(b\left(a-c\right)\)\(\ge\)0, c(b-a)\(\le\)0, a(c-b)\(\le\)0

\(a^3\ge b^3\ge c^3=>abc+a^3\ge abc+b^3\ge abc+c^3\)=>\(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}\le\dfrac{b\left(a-c\right)}{b\left(ac+b^2\right)}\)

=> VT -VP \(\le\) \(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}+\dfrac{c\left(b-a\right)}{b\left(ac+b^2\right)}+\dfrac{a\left(c-b\right)}{c\left(ab+c^2\right)}=\dfrac{ab-ac}{b\left(ac+b^2\right)}+\dfrac{ac-ab}{c\left(ab+c^2\right)}=\dfrac{a\left(b-c\right)}{b\left(ac+b^2\right)}-\dfrac{a\left(b-c\right)}{c\left(ab+c^2\right)}\)

mà \(\dfrac{1}{b\left(ac+b^2\right)}\le\dfrac{1}{c\left(ab+c^2\right)}\) nên VT-VP <0 đpcm

Ta viết bất đẳng thức đã cho lại thành

\(\sum\left[\dfrac{1}{c}-\dfrac{\left(a+b+2c\right)}{2\left(ab+c^2\right)}\right]\ge\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2\right)}{2\prod\left(ab+c^2\right)}\)

\(\Leftrightarrow\sum\dfrac{c\left(a^2+ab+b^2\right)\left(a-b\right)^2}{ab\left(a^2+bc\right)\left(b^2+ca\right)}\ge\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2\right)}{\prod\left(ab+c^2\right)}\)

Hay \(S_a\left(b-c\right)^2+S_b\left(c-a\right)^2+S_c\left(a-b\right)^2\ge\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2\right)}{\prod\left(ab+c^2\right)}\quad\left(1\right)\)

Vậy $VT\geq 0$ và $S_a+S_b\ge 0;S_b+S_c\ge 0.$ Nếu \(a\ge b\ge c\rightarrow VT\ge0\ge VP,\) ta chỉ xét \(a\le b\le c.\)

\(\left(1\right)\Leftrightarrow\left(S_a+S_b\right)\left(b-c\right)^2+\left(S_b+S_c\right)\left(a-b\right)^2\ge\left[\dfrac{\left(c-a\right)\left(a^2+b^2+c^2\right)}{\prod\left(ab+c^2\right)}-2S_b\right]\left(a-b\right)\left(b-c\right)\)

Đặt \(c=a+x+y,b=a+x\Rightarrow x=b-a;y=c-b\left(x,y\ge0\right)\) thay vào rút gọn các thứ là đpcm.

P/s: Cách này khá trâu nhưng chịu thôi, bài này mình nghĩ khá chặt.