Áp dụng bđt AM-GM ta có

\(abc\le\left(\frac{a+b+c}{3}\right)^3=1\)

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\sqrt[3]{\frac{1}{a^2b^2c^2}}\ge3\sqrt[3]{\frac{1}{a^3b^3c^3}}=\frac{3}{abc}\)

Ta chứng minh: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{abc}\)

\(\Leftrightarrow\frac{ab+bc+ca}{abc}\le\frac{3}{abc}\)

\(\Leftrightarrow ab+bc+ca\le3=\frac{\left(a+b+c\right)^2}{3}\)(luôn đúng)

Vậy bđt được chứng minh

Dấu "=" xảy ra khi a=b=c=1

Dòng thứ 3 của Linh bị ngược dấu rồi. 

Chứng minh các khác: 

Có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=3\)  (@)

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

Ta chứng minh: \(\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)(2)

<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)đúng theo (@) 

=> (2) đúng 

Từ (1) ; (2) => \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Dấu "=" xảy ra <=> a = b = c = 1.

Thank you các bạn, mk có cách khác hơi dài mới nghĩ ra nhờ mn xem hộ

Vì a,b,c>0 nên áp dụng bđt Cô-si ta có

\(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab};\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc};\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{a+b+c}{abc}\left(1\right)\)

Ta lại có: \(a^2+b^2\ge2ab;b^2+c^2\ge2bc;a^2+c^2\ge2ac\)\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

Cộng hai vế của bđt trên với 2ab+2bc+2ca ta được:\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)\(\Rightarrow3\left(a+b+c\right)\ge3\left(ab+bc+ca\right)\Rightarrow a+b+c\ge ab+bc+ca\)

\(\Rightarrow\frac{a+b+c}{abc}\ge\frac{ab+bc+ca}{abc}\Leftrightarrow\frac{a+b+c}{abc}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(2\right)\)

Từ (1) và (2) suy ra \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)(đpcm)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{1}{a^2}=\frac{1}{b^2};\frac{1}{b^2}=\frac{1}{c^2};\frac{1}{c^2}=\frac{1}{a^2}\\a^2=b^2;b^2=c^2;c^2=a^2\\a+b+c=3\end{cases}\Leftrightarrow a=b=c=1>0}\)

a/ \(\Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow x^2+y^2-2xy\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

b/ \(\frac{a}{a+b^2}=\frac{a}{a\left(a+b+c\right)+b^2}=\frac{a}{a^2+b^2+a\left(b+c\right)}\le\frac{a}{2ab+a\left(b+c\right)}=\frac{1}{b+b+b+c}\)

\(\Rightarrow\frac{a}{a+b^2}=\frac{1}{b+b+b+c}\le\frac{1}{16}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{16}\left(\frac{3}{b}+\frac{1}{c}\right)\)

Tương tự: \(\frac{b}{b+c^2}\le\frac{1}{16}\left(\frac{3}{c}+\frac{1}{a}\right)\) ; \(\frac{c}{c+a^2}\le\frac{1}{16}\left(\frac{3}{a}+\frac{1}{c}\right)\)

Cộng vế với vế:

\(VT\le\frac{1}{16}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

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

\(\sum\)\(\frac{a}{1+a^2}\)\(\le\)\(\sum\)\(\frac{a}{2a}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)

\(VT=\frac{a^2}{ab+ca}+\frac{b^2}{bc+ab}+\frac{c^2}{ca+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a+b+c\right)^2}{\frac{2}{3}\left(a+b+c\right)^2}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)

sao olm ko hiện \(\sum\) ra nhỉ ? thoi mk ghi lại v 

\(\frac{a}{1+a^2}\le\frac{a}{2a}=\frac{1}{2}\)

tương tự 2 cái kia cộng lại t có bđt cần cm 

\(\frac{a}{b+c}+\frac{b}{a+c}+\)\(\frac{c}{a+b}\ge\frac{3}{2}\)

Đặt b + c = x

      a + c = y

     a + b = z

Có: x + y - z = b + c + a + c - a - b = 2c

  \(\frac{x+y-z}{2}=c\)

Tương tự: \(\frac{x+z-y}{2}=b\)

     \(\frac{z+y-x}{2}=a\)

Khi đó:  = \(\frac{z+y-x}{2x}+\frac{x+z-y}{2y}\)\(+\frac{x+y-z}{2z}\)

= \(\frac{z+y}{2x}-\frac{x}{2x}\)\(+\frac{x+z}{2y}-\frac{y}{2y}+\)\(\frac{x+y}{2z}-\frac{z}{2z}\)

 = \(\frac{z+y}{2x}-\frac{1}{2}+\frac{x+z}{2y}-\frac{1}{2}\)\(+\frac{x+y}{2z}-\frac{1}{2}\)

 = \(\frac{z+y}{2x}+\frac{x+z}{2y}+\frac{x+y}{2z}\)\(-\frac{3}{2}\)

 = \(\frac{1}{2}.\left(\frac{z+y}{x}+\frac{x+z}{y}+\frac{x+y}{z}\right)\)\(-\frac{3}{2}\)

 = \(\frac{1}{2}.\)\(\left(\frac{z}{x}+\frac{y}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\right)\)\(-\frac{3}{2}\)

Ta có : \(\frac{z}{x}+\frac{x}{z}\ge2\)

            \(\frac{y}{x}+\frac{x}{y}\ge2\)

            \(\frac{y}{z}+\frac{z}{y}\ge2\)

\(\Rightarrow\)\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\)\(\frac{1}{2}.6-\frac{3}{2}\)

     \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\) ( đpcm )

Hình như đề sai, theo mik là nó lớn hơn bằng 3/2 nhé (ko biết đúng ko)

\(\frac{a}{b^2c+1}+\frac{b}{c^2a+1}+\frac{c}{a^2b+1}=\frac{a^2}{ab^2c+a}+\frac{b^2}{bc^2a+b}+\frac{c^2}{ca^2b+c}\)

Do a,b,c là 3 số thực dương nên áp dụng BĐT Cauchy Schwarz cho 3 phân số:

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

\(=\frac{\left(a+b+c\right)^2}{abc\left(a+b+c\right)+\left(a+b+c\right)}=\frac{9}{3abc+3}\)(Thay a+b+c=3)

Lại có: \(abc\le\frac{\left(a+b+c\right)^3}{27}=\frac{3^3}{27}=1\)(BĐT Cauchy cho 3 số)

\(\Rightarrow\frac{9}{3abc+3}\ge\frac{9}{6}=\frac{3}{2}\Rightarrow\frac{a^2}{ab^2c+a}+\frac{b^2}{bc^2a+b}+\frac{c^2}{ca^2b+c}\ge\frac{3}{2}\)

\(\Rightarrow\frac{a}{b^2c+1}+\frac{b}{c^2a+1}+\frac{c}{a^2b+1}\ge\frac{3}{2}.\)

\(\frac{1}{a+3b}+\frac{1}{a+b+2c}\ge\frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)

Tương tự: \(\frac{1}{b+3c}+\frac{1}{2a+b+c}\ge\frac{2}{a+b+2c}\) ; \(\frac{1}{c+3a}+\frac{1}{a+2b+c}\ge\frac{2}{2a+b+c}\)

Cộng vế với vế ta có đpcm

Ta chứng minh bổ đề:

\(\left(x+\frac{1}{x}\right)^2\ge\frac{260}{9}-\frac{160x}{3}\)

\(\Leftrightarrow\frac{9x^4+480x^3-242x^2+9}{9x^2}\ge0\)

\(\Leftrightarrow\frac{\left(3x-1\right)^2\left(x^2+54x+9\right)}{9x^2}\ge0\)(đúng)

Áp dụng vào bài toán ta được.

\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\)

\(\ge\frac{260}{9}-\frac{160a}{3}+\frac{260}{9}-\frac{160b}{3}+\frac{260}{9}-\frac{160c}{3}\)

\(=\frac{260}{3}-\frac{160}{3}\left(a+b+c\right)=\frac{260}{3}-\frac{160}{3}=\frac{100}{3}\)

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

áp dụng bunhia ta có:

\(\left(1+1+1\right)\left[\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\right]\ge\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right)^2\)

\(\ge\left(1+\frac{9}{a+b+c}\right)^2=100\)

\(\Rightarrow3\left[\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\right]\ge100\)

\(\Rightarrow\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\ge\frac{100}{3}\left(Q.E.D\right)\)

áp dụng bất đẳng thức Cauchy-schwaz

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{\left(1+1+1+1\right)^2}{a+b+c+d}\)=\(\frac{16}{a+b+c+d}\)(đpcm)

Dùng súng lục: "siêu tôc thần sầu" không đủ công lực tiếp nhận

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

 nhân phân phối bình thường ra thôi : \(t+\frac{1}{t}\ge2\)khi t>0 đẳng thức khi t=1

Áp vào trên => VT>=(1+1+1)+(2+2+2)=9

thay a+b+c=6 =>\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{6}=\frac{3}{2}\) =>dpcm

đẳng thúc khi t=1=> a/b=b/c=a/c=> a=b=c 

a+b+c=6=> a=b=c=2