1.

\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)

\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)

Dấu "=" khi \(a=b=c\)

2.

\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)

Dấu "=" khi \(a=b=c=d\)

Thục Trinh, tran nguyen bao quan, Phùng Tuệ Minh, Ribi Nkok Ngok, Lê Nguyễn Ngọc Nhi, Tạ Thị Diễm Quỳnh,

Nguyễn Huy Thắng, ?Amanda?, saint suppapong udomkaewkanjana

Help me!

a/ BĐT sai, cho \(a=b=c=2\) là thấy

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

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

Dấu "=" xảy ra khi \(a=b=c\)

c/ Tiếp tục sai nữa, vế phải là \(\frac{3}{2}\) chứ ko phải \(2\), và hy vọng rằng a;b;c dương

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

\(VT\ge\frac{9}{3abc+3}\ge\frac{9}{\frac{3\left(a+b+c\right)^3}{27}+3}=\frac{9}{\frac{3.3^3}{27}+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có:

\(a^3+b^3+b^3\ge3ab^2\) ; \(b^3+c^3+c^3\ge3bc^2\) ; \(c^3+a^3+a^3\ge3ca^2\)

Cộng vế với vế \(\Rightarrow a^3+b^3+c^3\ge ab^2+bc^2+ca^2\)

\(\frac{a^5}{b^2}+\frac{b^5}{c^2}+\frac{c^5}{a^2}=\frac{a^6}{ab^2}+\frac{b^6}{bc^2}+\frac{c^6}{ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab^2+bc^2+ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{a^3+b^3+c^3}=a^3+b^3+c^3\)

Câu 2)

Ta có \(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{3}\)

\(\Rightarrow\frac{b+1+a+1}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)

Ta có \(a+b=1\)

\(\Rightarrow\frac{3}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)

\(\Rightarrow\frac{3}{\left(a+1\right)b+a+1}\ge\frac{4}{3}\)

\(\Rightarrow\frac{3}{ab+b+a+1}\ge\frac{4}{3}\)

Ta có \(a+b=1\)

\(\Rightarrow\frac{3}{ab+2}\ge\frac{4}{3}\)

\(\Leftrightarrow9\ge4\left(ab+2\right)\)

\(\Rightarrow9\ge4ab+8\)

\(\Rightarrow1\ge4ab\)

Do \(a+b=1\Rightarrow\left(a+b\right)^2=1\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

\(\Rightarrow a^2+2ab+b^2\ge4ab\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow\left(a-b\right)^2\ge0\) (đpcm )

Câu 3)

Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)

Mà \(a+b+c=1\)

\(\Rightarrow\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\ge9\)

\(\Rightarrow a+b+c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

Áp dụng bất đẳng thức Cô-si

\(\Rightarrow\left\{\begin{matrix}a+b+c\ge3\sqrt[3]{abc}\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\end{matrix}\right.\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc}\sqrt[3]{\frac{1}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\sqrt[3]{\frac{abc}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (điều này luôn luôn đúng)

\(\Rightarrow\) ĐPCM

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

Dấu "=" xảy ra khi \(a=b=c\)

Hoặc bạn dùng AM-GM kiểu:

\(\frac{a^3}{b+c}+a\left(b+c\right)\ge2a^2\)

Làm tương tự với 2 cái sau và cộng lại

Ngoài ra có cách dùng AM-GM cho 3 số như sau:

Ta có: \(\frac{a^3}{b+c}+\frac{a^3}{b+c}+\frac{\left(b+c\right)^2}{8}\ge\frac{3}{2}a^2\)

Tương tự rồi cộng lại:

\(2VT\ge\frac{3}{2}\left(a^2+b^2+c^2\right)+\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{8}\)

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

Sorry, tới đây em bí rồi ạ:v

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

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