\(\frac{a^3}{\left(b+2\right)^2}+\frac{b+2}{27}+\frac{b+2}{27}\ge3\sqrt[3]{\frac{a^3\left(b+2\right)^2}{27^2.\left(b+2\right)^2}}=\frac{a}{3}\)

Tương tự: \(\frac{b^3}{\left(c+2\right)^2}+\frac{c+2}{27}+\frac{c+2}{27}\ge\frac{b}{3}\) ; \(\frac{c^3}{\left(a+2\right)^2}+\frac{a+2}{27}+\frac{a+2}{27}\ge\frac{c}{3}\)

Cộng vế với vế:

\(VT+\frac{2\left(a+b+c\right)+12}{27}\ge\frac{a+b+c}{3}\)

\(\Leftrightarrow VT+\frac{2}{3}\ge1\Leftrightarrow VT\ge\frac{1}{3}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

b/

\(a^3+a^3+1\ge3\sqrt[3]{a^6}=3a^2\)

Tương tự: \(2b^3+1\ge3b^2\) ; \(2c^3+1\ge3c^2\)

Cộng vế với vế:

\(2\left(a^3+b^3+c^3\right)\ge3\left(a^2+b^2+c^2\right)-3\)

Mặt khác ta lại có:

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

\(\Rightarrow2\left(a^3+b^3+c^3\right)\ge2\left(a^2+b^2+c^2\right)+\left(a^2+b^2+c^2\right)-3\ge2\left(a^2+b^2+c^2\right)+3-3\)

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

Dấu "=" xảy ra khi \(a=b=c=1\)

Dạ em cảm ơn ạ

Sửa đề: \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3}{a+b+c}\ge4\)

\(\Leftrightarrow\frac{a^2c+b^2a+c^2b}{abc}+\frac{3}{a+b+c}\ge4\)

\(\Leftrightarrow P=a^2c+b^2a+c^2b+\frac{3}{a+b+c}\ge4\)

Ta có:

\(a^2c+a^2c+b^2a\ge3\sqrt[3]{a^3.\left(abc\right)^2}=3a\)

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

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

Cộng vế với vế: \(a^2c+b^2a+c^2b\ge a+b+c\)

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

\(\Rightarrow P\ge2\sqrt{\frac{3\left(a+b+c\right)}{3\left(a+b+c\right)}}+\frac{2}{3}.3\sqrt[3]{abc}=4\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có: \(\frac{a^2}{b}+\frac{b^2}{a}+7\left(a+b\right)\ge8\sqrt{2\left(a^2+b^2\right)}\)

\(\Leftrightarrow a^3+b^3+7ab\left(a+b\right)\ge8ab\sqrt{2\left(a^2+b^2\right)}\)

Ta có: \(VP=8\sqrt{ab}\sqrt{\left(a^2+b^2\right)\cdot2ab}\le^{am-gm}4\sqrt{ab}\left(a+b\right)^2\)

\(VT=\left(a+b\right)\left[\left(a+b\right)^2+4ab\right]\ge^{am-gm}\left(a+b\right)4\sqrt{ab}\left(a+b\right)\ge VP\)

=> ĐPCM

Vì a+b+c=0=>(a+b)=-c. Tương tự:(b+c)=-a;(a+c)=-b.

Ta có A=:\(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)

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

^{\(=\frac{a^2}{\left(a-b\right).\left(-c\right)-c^2}+tươngtự\)}

^{\(=\frac{a^2}{-ca+bc-c^2}\)+ tương tự}

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

^{\(=\frac{a^2}{c\left(b-\left(c+a\right)\right)}\)}+ tương tự nha 

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

Sau đó ta có :\(\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2bc}\)

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

\(=\frac{\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b\right)}{2abc}\)=\(\frac{0-0-3ab\left(-c\right)}{2abc}\)(do a+b+c=0)

=\(\frac{3abc}{2abc}=\frac{3}{2}\)Ok r bạn

Ta có 

\(\frac{a^2}{a+b^2}=\frac{a^2+ab^2-ab^2}{a+b^2}=a-\frac{ab^2}{a+b^2}\ge a-\frac{b\sqrt{a}}{2}\ge a-\frac{1}{4}b\left(a+1\right)\)

Khi đó 

\(A\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{4}\left(ab+bc+ac\right)\)

Mà \(ab+bc+ac\le\frac{1}{3}\left(a+b+c\right)^2=3\)

=> \(A\ge\frac{9}{4}-\frac{3}{4}=\frac{3}{2}\)( ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

\(a-\frac{ab^2}{a+b^2}\ge a-\frac{b\sqrt{a}}{2}\)

Do \(a+b^2\ge2b\sqrt{a}\)

\(a-\frac{ab^2}{a+b^2}\ge a-\frac{b\sqrt{a}}{2}\ge a-\frac{1}{4}b\left(a+1\right)\)

Do \(\sqrt{a}\le\frac{a+1}{2}\)

câu a,mình ko biết nhưng câu b bạn cộng 1+b cho số hạng đầu áp dụng cô si,các số hạng khác tương tự rồi cộng vế theo vế,ta có điều phải c/m

Bạn nói rõ hơn được không???

Để chừng nào t làm được câu 1 thì t giải giúp cho 1 lần luôn

ap dung bdt am gm

\(\sqrt{1+8a^3}=\sqrt{\left(1+2a\right)\left(4a^2-4a+1\right)}\)\(\le\frac{1+2a+4a^2-2a+1}{2}=\frac{4a^2+2}{2}=2a^2+1\)

\(\Rightarrow\frac{1}{\sqrt{1+8a^3}}\ge\frac{1}{2a^2+1}\)

tuongtu ta cung co \(\frac{1}{\sqrt{1+8b^3}}\ge\frac{1}{2b^2+1};\frac{1}{\sqrt{1+8c^3}}\ge\frac{1}{2c^2+1}\)

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

tiep tuc ap dung bat cauchy-schwarz dang engel ta co

\(VT\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\ge\frac{\left(1+1+1\right)^2}{2\left(a^2+b^2+c^2\right)+3}=\frac{3^2}{6+3}=1\)(dpcm)

dau = xay ra \(\Leftrightarrow a=b=c=1\)