dạng này thì chỉ có quy đồng thôi nhé mặc dù quy đồng chưa ra

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

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

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

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

\(=2+ab+bc+ca=VP\) (Do a² + b² + c² = 1) => ĐPCM.

Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{\sqrt{3}}.\)

chăc là .............................. điền đi sẽ biếc a you ok ?

dễmaf

thay c=c.1=c(a+b+c)

=> ab+c=(c+a)(c+b)

lm tt cuối cùng sẽ ra

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

Áp dụng bđt Cauchy dạng phân thức 

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

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

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

\(\Leftrightarrow VT\ge\frac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)

Dấu " = " xảy ra khi \(a=b=c\)

Chúc bạn học tốt !!!

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

Tương tự: \(\sqrt{\frac{bc}{a+bc}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\) ; \(\sqrt{\frac{ca}{b+ca}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{a+b}\right)\)

Cộng vế với vế: \(VT\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{a}{a+c}+\frac{c}{a+c}\right)=\frac{3}{2}\)

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

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)

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

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

Áp dụng BĐT Cauchy , có :

\(\frac{a^3+b^3}{ab}+\frac{b^3+c^3}{bc}+\frac{c^3+a^3}{ca}\ge\frac{2\sqrt{a^3.b^3}}{ab}+\frac{2\sqrt{b^3.c^3}}{bc}+\frac{2\sqrt{c^3.a^3}}{ca}\)

\(\Leftrightarrow...........\ge2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\)

Lại có :

\(2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\ge2a+2b+2c\)

\(\Leftrightarrow\left(a-2\sqrt{ab}+b\right)+\left(b-2\sqrt{bc}+c\right)+\left(c-2\sqrt{ca}+c\right)\ge0\)

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

Vậy \(\frac{a^3+b^3}{ab}+\frac{b^3+c^3}{bc}+\frac{c^3+a^3}{ca}\ge2a+2b+2c=2\left(a+b+c\right)\)

Ta có BĐT \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)

\(\Rightarrow\frac{a^3+b^3}{ab}\ge\frac{ab\left(a+b\right)}{ab}=a+b\)

Tương tự cũng có 2 BĐT:

\(\frac{b^3+c^3}{bc}\ge b+c;\frac{c^3+a^3}{ca}\ge c+a\)

Cộng theo vế được ĐPCM

Khi a=b=c

Ta có BĐT \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)

\(\Rightarrow\frac{a^3+b^3}{ab}\ge\frac{ab\left(a+b\right)}{ab}=a+b\)

Tương tự cũng có 2 BĐT:

\(\frac{b^3+c^3}{bc}\ge b+c;\frac{c^3+a^3}{ca}\ge c+a\)

Cộng theo vế được ĐPCM

Khi a=b=c