3/ Áp dụng bất đẳng thức AM-GM, ta có :

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

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)

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

Cộng 3 vế của BĐT trên ta có :

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

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

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)

Do đó:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Bài 2:

Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)

\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)

\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\text{VT}=\frac{1}{1+ab}+\frac{a^2}{a+ab}+\frac{b^2}{b+ab}\geq \frac{(1+a+b)^2}{1+ab+a+ab+b+ab}\)

\(\Leftrightarrow \text{VT}\geq \frac{(a+b+1)^2}{a+b+1+3ab}\)

\(\Leftrightarrow \text{VT}\geq \frac{(a+b+1)^2}{a+b+1+3(3-a-b)}=\frac{(a+b+1)^2}{10-2(a+b)}\)

Theo giả thiết:

\(3=a+b+ab\Leftrightarrow 4=a+b+ab+1=(a+1)(b+1)\)

\(\leq \left (\frac{a+b+2}{2}\right)^2\) (theo BĐT AM-GM)

suy ra \(a+b+2\geq 4\Leftrightarrow a+b\geq 2\) (với \(a,b>0\) )

Do đó: \((a+b+1)^2\geq 9\) (1)

\(10-2(a+b)\leq 10-2.3=4; 10-2(a+b)=4+2ab>0\)

\(\Rightarrow \frac{1}{10-2(a+b)}\geq \frac{1}{6}\) (2)

Từ \((1);(2)\Rightarrow A\geq \frac{(a+b+1)^2}{10-2(a+b)}\geq \frac{9}{6}=\frac{3}{2}\)

Ta có đpcm

Dấu bằng xảy ra khi \(a=b=1\)

Do \(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)

\(VT=\dfrac{xz}{y\left(x+z\right)}+\dfrac{xy}{z\left(x+y\right)}+\dfrac{yz}{x\left(y+z\right)}=\dfrac{\left(xz\right)^2}{xyz\left(x+z\right)}+\dfrac{\left(xy\right)^2}{xyz\left(x+y\right)}+\dfrac{\left(yz\right)^2}{xyz\left(y+z\right)}\)

\(VT\ge\dfrac{\left(xy+yz+zx\right)^2}{2xyz\left(x+y+z\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{2xyz\left(x+y+z\right)}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=1\)

bạn làm được bài nảy chưa ? chỉ mình với

Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)

\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)

CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)

Cộng vế theo vế:

\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

Lời giải:

a)

Sử dụng pp biến đổi tương đương:

\(\frac{1}{a^2+1}+\frac{1}{b^2+1}\geq \frac{2}{ab+1}\Leftrightarrow \frac{a^2+b^2+2}{(a^2+1)(b^2+1)}\geq \frac{2}{ab+1}\)

\(\Leftrightarrow (ab+1)(a^2+b^2+2)\geq 2(a^2b^2+a^2+b^2+1)\)

\(\Leftrightarrow ab(a^2+b^2)+2ab\geq 2a^2b^2+a^2+b^2\)

\(\Leftrightarrow ab(a^2+b^2-2ab)-(a^2+b^2-2ab)\geq 0\)

\(\Leftrightarrow ab(a-b)^2-(a-b)^2\geq 0\)

\(\Leftrightarrow (ab-1)(a-b)^2\geq 0\) (luôn đúng với mọi $ab\geq 1$)

Ta có đpcm.

b) Áp dụng công thức của phần a ta có:

\(\frac{1}{a^4+1}+\frac{1}{b^4+1}\geq \frac{2}{1+(ab)^2}\)

Tiếp tục áp dụng công thức phần a: \(\frac{1}{1+(ab)^2}+\frac{1}{1+b^4}\geq \frac{2}{1+ab^3}\)

Do đó:

\(\frac{1}{a^4+1}+\frac{3}{b^4+1}\geq \frac{4}{1+ab^3}\)

Hoàn toàn tương tự: \(\frac{1}{b^4+1}+\frac{3}{c^4+1}\geq \frac{4}{1+bc^3}; \frac{1}{c^4+1}+\frac{3}{a^4+1}\geq \frac{4}{1+ca^3}\)

Cộng theo vế các BĐT trên thu được:

\(4\left(\frac{1}{a^4+1}+\frac{1}{b^4+1}+\frac{1}{c^4+1}\right)\geq 4\left(\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\right)\)

\(\Leftrightarrow \frac{1}{a^4+1}+\frac{1}{b^4+1}+\frac{1}{c^4+1}\geq \frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)

Ta có đpcm

Dấu bằng xảy ra khi $a=b=c=1$

\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)

Xét \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)

\(\Rightarrow VT=\dfrac{a^3}{a^2+ab+bc+ac}+\dfrac{b^3}{b^2+ab+bc+ac}+\dfrac{c^3}{c^2+ab+bc+ac}\)

\(\Leftrightarrow VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\)

Áp dụng bđt Cauchy ta có :

\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)

Thiết lập tương tự và thu lại ta có :

\(VT+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)

\(\Rightarrow VT\ge\dfrac{3}{4}\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{4}\left(đpcm\right)\)

Dấu '' = '' xảy ra khi \(a=b=c=3\)

sai đề: cho \(a,b\ge1\) mới chuẩn

Ta có :

\(\left(a-1\right)^2\ge0\)

\(\Rightarrow a^2+1\ge2a\)

\(\Rightarrow2a-1\le a^2\)

\(\Rightarrow\dfrac{a}{2a-1}\ge\dfrac{a}{a^2}\)

\(\Rightarrow\dfrac{a}{2a-1}\ge\dfrac{1}{a}\)

Tương tự ta có :

\(\dfrac{b}{2b-1}\ge\dfrac{1}{b}\)

Do đó : \(\dfrac{a}{2a-1}+\dfrac{b}{2b-1}\ge\dfrac{1}{a}+\dfrac{1}{b}\)

*) Chứng minh bổ đề : \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\left(x,y>0\right)\)

Ta có :

\(\left(x-y\right)^2\ge0\)

\(\Rightarrow x^2+y^2\ge2xy\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\)

\(\Rightarrow\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}\ge\dfrac{4xy}{xy\left(x+y\right)}\left(x,y>0\right)\)

\(\Rightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\)

\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

*) Áp dụng bổ đè trên ta có:

\(\dfrac{a}{2a-1}+\dfrac{b}{2b-1}\ge\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) (1)

Ta có :

\(a+b-\left(1+ab\right)\)

\(=\left(a-ab\right)+\left(b-1\right)\)

\(=a\left(1-b\right)+\left(b-1\right)\)

\(=\left(1-b\right)\left(a-1\right)\)

Vì \(a,b\ge1\Rightarrow\left\{{}\begin{matrix}a-1\ge0\\1-b\le0\end{matrix}\right.\)

\(\Rightarrow\left(1-b\right)\left(a-1\right)\le0\)

\(\Rightarrow a+b-\left(1+ab\right)\le0\)

\(\Rightarrow a+b\le1+ab\)

\(\Rightarrow\dfrac{4}{a+b}\ge\dfrac{4}{1+ab}\) (2)

Từ (1) và (2) ta được:

\(\dfrac{a}{2a-1}+\dfrac{b}{2b-1}\ge\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\ge\dfrac{4}{1+ab}\)

\(\Rightarrow\dfrac{a}{2a-1}+\dfrac{b}{2b-1}\ge\dfrac{4}{1+ab}\)

\(\rightarrowđpcm\)

Áp dụng BĐT BSC:

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

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

\(\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{1}{2}\)

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