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biến đổi tương đương đi, nhân tung ngoặc ra

\(\left(a+b\right)\left(a^4+b^4\right)\ge\left(a^2+b^2\right)\left(a^3+b^3\right)\)

\(\Leftrightarrow a^5+ab^4+a^4b+b^5\ge a^5+a^2b^3+a^3b^2+b^5\)

\(\Leftrightarrow ab^4+a^4b-a^2b^3-a^3b^2\ge0\)

\(\Leftrightarrow ab\left(a^3+b^3-ab^2-a^2b\right)\ge0\)

\(\Leftrightarrow a^3+b^3-ab^2-a^2b\ge0\)(Do ab > 0)

\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)Luôn đúng do a,b dương

Dấu "='' khi a = b

Xét \(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)

Tương tự, ta được: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)

Cộng theo vế của 3 đẳng thức trên, ta được: \(\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)\(-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)=0\)

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

Ta đi chứng minh BĐT phụ sau: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)(*)

Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\ge0\)*đúng*

\(\Rightarrow2LHS=\Sigma_{cyc}\frac{a^3+b^3}{a^2+ab+b^2}=\Sigma_{cyc}\text{ }\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\)\(\ge\Sigma_{cyc}\text{ }\frac{\frac{1}{3}\left(a+b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=\frac{1}{3}\text{​​}\Sigma_{cyc}\left[\left(a+b\right)\right]=\frac{2\left(a+b+c\right)}{3}\)

\(\Rightarrow LHS\ge\frac{a+b+c}{3}=RHS\)(Q.E.D)

Đẳng thức xảy ra khi a = b = c

P/S: Có thể dùng BĐT phụ ở câu 3a để chứng minhxD:

1) ta chứng minh được \(\Sigma\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}=\Sigma\frac{b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)

\(VT=\frac{1}{2}\Sigma\frac{a^4+b^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{1}{4}\Sigma\frac{a^2+b^2}{a+b}\ge\frac{1}{8}\Sigma\left(a+b\right)=\frac{a+b+c+d}{4}\)

bài 2 xem có ghi nhầm ko

3a biến đổi tí là xong

b tuong tự bài 1 

\(BĐT\Leftrightarrow\left(a+b+c\right)\left(\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\right)\ge\frac{9}{4}\)

Áp dụng BĐT Bunhi kết hợp với Nesbit :

\(VT=\left(\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2\right)\left[\left(\frac{\sqrt{a}}{b+c}\right)^2+\left(\frac{\sqrt{b}}{c+a}\right)^2+\left(\frac{\sqrt{c}}{a+b}\right)^2\right]\ge\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\ge\left(\frac{3}{2}\right)^2=\frac{9}{4}\)

Vậy BĐT đc chứng minh . Dấu bằng xảy ra khi \(a=b=c\)

Bất đẳng thức

<=> \(\frac{a\left(a+b+c\right)}{\left(b+c\right)^2}+\frac{b\left(a+b+c\right)}{\left(c+a\right)^2}+\frac{c\left(a+b+c\right)}{\left(a+b\right)^2}\ge\frac{9}{4}\)

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

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

lại có:

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

\(\ge\left(a+b+c\right).\frac{9}{2\left(a+b+c\right)}-3=\frac{3}{2}\)

=> VT\(\ge\frac{1}{3}.\left(\frac{3}{2}\right)^2+\frac{3}{2}=\frac{9}{4}\)

Dấu "=" xảy ra <=> a = b = c.

Hoặc em có thể áp dụng Bunhia

bất đẳng thức 

<=> \(\left(a+b+c\right)\left(\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\right)\ge\frac{9}{4}\)

VT\(\ge\left(\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\right)^2\ge\left(\frac{3}{2}\right)^2=\frac{9}{4}\)

Cảm ơn chị