theo bài ra ta có: \(c^2+2ab-2bc-2ca=0.\)

\(\Rightarrow2\left(c^2+ab-bc-ca\right)=c^2\)

\(\Rightarrow2\left(a-c\right)\left(b-c\right)=c^2\)

Mặt khác: \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{2a^2-2ac+c^2}{2b^2-2bc+c^2}=\frac{2a\left(a-c\right)+2\left(a-c\right)\left(b-c\right)}{2b\left(b-c\right)+2\left(a-c\right)\left(b-c\right)}\)

                                                                               \(=\frac{\left(a-c\right)\left(a+b-c\right)}{\left(b-c\right)\left(b+a-c\right)}=\frac{a-c}{b-c}\) => đpcm

Áp dụng BĐT Cauchy-SChwarz ta có:

\(VT=\frac{a^4}{a^2+2a^2bc}+\frac{b^4}{b^2+2ab^2c}+\frac{c^4}{c^2+2abc^2}\)

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

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

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

\(\ge\frac{1^2}{1+2\cdot\frac{1^2}{3}}=\frac{3}{5}=VP\)

Dấu "=" bạn tự nghiên cứu nhé :D

DẤU BẰNG XẢY RA\(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\) CÁI NÀY LÀ ĐIỂM RƠI NHÉ.

Cách : AM - GM :

\(VT=3-\left(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}\right)\left(1\right)\)

Áp dụng BĐT AM - GM :

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

\(\le\frac{2ab^2}{3\sqrt[3]{a^2b^4}}+\frac{2bc^2}{3\sqrt[3]{b^2c^4}}+\frac{2ca^2}{3\sqrt[3]{c^aa^4}}=\frac{2}{3}\left(\sqrt[3]{ab^2}+\sqrt[3]{bc^2}+\sqrt[3]{ca^2}\right)\)

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

Từ (1) và (2) \(\Rightarrow VT\ge3-2=1\left(đpcm\right)\)

Ta có: \(\frac{a}{c}=\frac{a^2+b^2}{c^2+b^2}\)\(\Leftrightarrow a\left(c^2+b^2\right)=c\left(a^2+b^2\right)\)\(\Leftrightarrow ac^2+ab^2=a^2c+b^2c\Leftrightarrow ac\left(c-a\right)-b^2\left(c-a\right)=0\)

\(\Leftrightarrow\left(c-a\right)\left(ac-b^2\right)=0\)

Vì \(a\ne c\)nên \(c-a\ne0\)

Do đó \(ac-b^2=0\Leftrightarrow ac=b^2\Rightarrow\sqrt{ac}=b\)

Giả sử \(a^2+b^2+c^2\)là số nguyên tố

Ta có \(a^2+b^2+c^2=a^2+ac+c^2=\left(a+c\right)^2-ac=\left(a+c\right)^2-b^2\)\(=\left(a-b+c\right)\left(a+b+c\right)\)

\(=\left[\left(\sqrt{a}\right)^2-2\sqrt{ac}+\left(\sqrt{c}\right)^2+\sqrt{ac}\right]\left[\left(\sqrt{a}\right)^2-2\sqrt{ac}+\left(\sqrt{c}\right)^2+3\sqrt{ac}\right]\)

\(\left[\left(\sqrt{a}-\sqrt{c}\right)^2+\sqrt{ac}\right]\left[\left(\sqrt{a}-\sqrt{c}\right)^2+3\sqrt{ac}\right]\)

Vì \(a^2+b^2+c^2\)là số nguyên tố nên có một ước số là 1

Mà \(\left(\sqrt{a}-\sqrt{c}\right)^2+\sqrt{ac}< \left(\sqrt{a}-\sqrt{c}\right)^2+3\sqrt{ac}\)

nên \(\left(\sqrt{a}-\sqrt{c}\right)^2+\sqrt{ac}=1\Leftrightarrow\left(\sqrt{a}-\sqrt{c}\right)^2=1-\sqrt{ac}\)

Vì \(a\ne c\Rightarrow\sqrt{a}\ne\sqrt{c}\Rightarrow\sqrt{a}-\sqrt{c}\ne0\)\(\Rightarrow\left(\sqrt{a}-\sqrt{c}\right)^2>0\)

Do đó \(1-\sqrt{ac}>0\Rightarrow\sqrt{ac}< 1\Rightarrow ac< 1\)(1)

Mà \(a^2+b^2>0\)và \(c^2+b^2>0\)nên \(\frac{a^2+b^2}{c^2+b^2}>0\Rightarrow\frac{a}{c}>0\Rightarrow\)a, c cùng dấu \(\Rightarrow ac>0\)(2)

Từ (1), (2) suy ra \(0< ac< 1\)

Mà a,c là số nguyên nên ac là số nguyên 

Do đó không có giá trị a,c thỏa mãn

suy ra điều giả sử sai

Vậy \(a^2+b^2+c^2\) không thể là số nguyên tố

tự giải vl

Đặt \(P=\frac{a}{a+2bc}+\frac{b}{b+2ca}+\frac{c}{c+2ab}\)

\(\Leftrightarrow P=\frac{a^2}{a^2+2bca}+\frac{b^2}{b^2+2cab}+\frac{c^2}{c^2+2abc}\)

Áp dụng BĐT Cauchy-schwarz ta có: ( link c/m Cauchy-schwarz: Xem câu hỏi )

\(P\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+6abc}=\frac{9}{a^2+b^2+c^2+6abc}\)( \(a+b+c=3\))

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

Ta có: \(a+b+c=3\)

Áp dụng BĐT AM-GM ta có:

\(a+b+c\ge3\sqrt[3]{abc}\)

\(\Leftrightarrow3\ge3\sqrt[3]{abc}\)

\(\Leftrightarrow1\ge\sqrt[3]{abc}\)

\(\Leftrightarrow1\ge abc\)

\(\Leftrightarrow a^2b^2c^2\ge a^3b^3c^3\)

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

Áp dụng BĐT AM-GM ta có:

\(ab+bc+ca\ge3.\sqrt[3]{a^2b^2c^2}\ge3.\sqrt[3]{a^3b^3c^3}=3abc\)

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

\(\Rightarrow P\ge\frac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{9}{\left(a+b+c\right)^2}=\frac{9}{9}=1\)

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

                                đpcm

với mọi x,y,z >0 ta có: \(x+y+z\ge3\sqrt[3]{xyz};\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{xyz}}\)

\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)

\(\Rightarrow\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

đẳng thức xảy ra khi x=y=z

ta có: \(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)

đẳng thức xảy ra khi a=b

tương tự: \(\frac{1}{\sqrt{5b^2+2ab+2b^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)

đẳng thức xảy ra khi b=c

\(\frac{1}{\sqrt{5c^2+2bc+2c^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)

đẳng thức xảy ra khi c=a

Vậy \(\frac{1}{\sqrt{5a^2+2ca+2a^2}}+\frac{1}{\sqrt{5b^2+2bc+2c^2}}+\frac{1}{\sqrt{5c^2+2ac+2a^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)

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

đẳng thức xảy ra khi a=b=c=\(\frac{3}{2}\)

Tham khảo bài của mình

1. BĐT ban đầu

<=> \(\left(\frac{1}{3}-\frac{b}{a+3b}\right)+\left(\frac{1}{3}-\frac{c}{b+3c}\right)+\left(\frac{1}{3}-\frac{a}{c+3a}\right)\ge\frac{1}{4}\)

<=>\(\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)

<=> \(\frac{a^2}{a^2+3ab}+\frac{b^2}{b^2+3bc}+\frac{c^2}{c^2+3ac}\ge\frac{3}{4}\)

Áp dụng BĐT buniacoxki dang phân thức 

=> BĐT cần CM

<=> \(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ac\right)}\ge\frac{3}{4}\)

<=> \(a^2+b^2+c^2\ge ab+bc+ac\)luôn đúng 

=> BĐT được CM

2) \(a+b+c\le ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(\left(a+b+c\right)^2-3\left(a+b+c\right)\ge0\)

\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a+b+c-3\right)\ge0\)\(\Leftrightarrow\)\(a+b+c\ge3\)

ko mất tính tổng quát giả sử \(a\ge b\ge c\)

Có: \(3\le a+b+c\le ab+bc+ca\le3a^2\)\(\Leftrightarrow\)\(3a^2\ge3\)\(\Leftrightarrow\)\(a\ge1\)

=> \(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le\frac{3}{1+2a}\le1\)

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