Ta có:

\((p-a)(p-b) \leq \frac{(p-a+p-b)^2}{4}=\frac{c^2}{4}\) tương tự rồi nhân lại, ta có đpcm.

Ta có bất đẳng thức phụ sau:

 \(\left(a+b-c\right)\left(c+a-b\right)\left(b+c-a\right)\le abc\)  \(\left(\text{*}\right)\) với  \(a,b,c\)  là độ dài ba cạnh của một tam giác  \(\left(a,b,c>0\right)\)

Thật vậy,  áp dụng bất đẳng thức AM-GM cho các cặp số dương:

\(\left(a+b-c\right)+\left(c+a-b\right)\ge2\sqrt{\left(a+b-c\right)\left(c+a-b\right)}\)  

\(\Rightarrow\)  \(2a\ge2\sqrt{\left(a+b-c\right)\left(c+a-b\right)}\)

\(\Rightarrow\)  \(a^2\ge\left(a+b-c\right)\left(c+a-b\right)\)  \(\left(1\right)\)

Tương tự áp dụng bất đẳng trên, ta cũng được:

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

Từ  \(\left(1\right);\)  \(\left(2\right)\)  và  \(\left(3\right)\)  \(\left(abc\right)^2\ge\left[\left(a+b-c\right)\left(c+a-b\right)\left(b+c-a\right)\right]^2\)

                                  \(\Rightarrow\)  \(\left(a+b-c\right)\left(c+a-b\right)\left(b+c-a\right)\le abc\)  

Dấu  \(''=''\)  xảy ra  \(\Leftrightarrow\)  \(a=b=c\)  \(\Leftrightarrow\)  tam giác đó là tam giác đều

 Ta dễ dàng nhận thấy:

 \(p-a=\frac{a+b+c}{2}-a=\frac{a+b+c-2a}{2}=\frac{b+c-a}{2}\)

 \(p-b=\frac{a+b+c}{2}-b=\frac{a+b+c-2b}{2}=\frac{c+a-b}{2}\)

  \(p-c=\frac{a+b+c}{2}-c=\frac{a+b+c-2c}{2}=\frac{a+b-c}{2}\)

Do đó,  \(\left(p-a\right)\left(p-b\right)\left(p-c\right)=\frac{b+c-a}{2}.\frac{c+a-b}{2}.\frac{a+b-c}{2}=\frac{\left(a+b-c\right)\left(c+a-b\right)\left(b+c-a\right)}{8}\)  \(\left(\text{**}\right)\)  

Từ  \(\left(\text{*}\right)\)  và  \(\left(\text{**}\right)\)  ta có:  \(\left(p-a\right)\left(p-b\right)\left(p-c\right)\le\frac{abc}{8}\)  

Dấu bằng xảy ra  khi và chỉ khi  \(a=b=c\)  \(\Leftrightarrow\)  tam giác đó làm tam giác đều

Giúp em cái

Ta có:

\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge\frac{3a}{4}\)

\(\Rightarrow\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge\frac{4a-b-c}{8}\left(1\right)\)

Tương tự ta có: 

\(\hept{\begin{cases}\frac{b^3}{\left(b+a\right)\left(b+c\right)}\ge\frac{4b-a-c}{8}\left(2\right)\\\frac{c^3}{\left(c+a\right)\left(c+b\right)}\ge\frac{4c-a-b}{8}\left(3\right)\end{cases}}\)

Cộng (1), (2), (3) vế theo vế được

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

từ dòng 1 xuống dòng 2 mình không hiểu lắm

Chuyển vế mà quên xóa ấy mà. 

\(3=a+b+c\ge3\sqrt[3]{abc}\)\(\Leftrightarrow\)\(abc\le1\)

\(VT=\frac{a^3\left(a+1\right)+b^3\left(b+1\right)+c^3\left(c+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=\frac{a^4+b^4+c^4+a^3+b^3+c^3}{a+b+c+ab+bc+ca+abc+1}\)

\(\ge\frac{\frac{\left(a^2+b^2+c^2\right)^2}{3}+\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c}}{\frac{\left(a+b+c\right)^2}{3}+5}=\frac{\frac{\frac{\left(a+b+c\right)^4}{9}}{3}+\frac{\frac{\left(a+b+c\right)^4}{9}}{3}}{8}\)

\(=\frac{\frac{\frac{3^4}{9}}{3}}{4}=\frac{3}{4}\)

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

đề viết gì thế bạn ?

>= 3/4

\(a\left(a+b\right)\left(a+c\right)=b\left(b+c\right)\left(b+a\right)\)

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

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

\(\Leftrightarrow\left(a+b\right)\left[\left(a+b\right)\left(a-b\right)+c\left(a-b\right)\right]=0\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)\left(a+b+c\right)=0\)

\(\Leftrightarrow a+b=0\)hoặc \(a-b=0\)hoặc \(a+b+c=0\)

\(\Leftrightarrow a=b\)(Không thỏa điều kiện) hoặc a=-b (Không thỏa điều kiện) hoặc a+b+c=0

<=> a+b+c=0 (đpcm)

a,

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=2\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

\(\Leftrightarrow a=b=c\)

b,

\(a+b+c=2p\Leftrightarrow p=\dfrac{a+b+c}{2}\)

\(\Leftrightarrow\left(p-a\right)^2+\left(p-b\right)^2+\left(p-c\right)^2=3p^2-2pa-2pb-2pc+a^2+b^2+c^2\)

\(=3\left(\dfrac{a+b+c}{2}\right)^2-2\cdot\dfrac{a+b+c}{2}\cdot a-2\cdot\dfrac{a+b+c}{2}\cdot b-2\cdot\dfrac{a+b+c}{2}\cdot c+a^2+b^2+c^2\)

\(=3p^2-\left(a+b+c\right)^2+a^2+b^2+c^2=3p^2-4p^2+a^2+b^2+c^2=a^2+b^2+c^2-p^2\)

Áp dụng BĐT Cosi:

\(\frac{a^4}{\left(a+2\right)\left(b+2\right)}+\frac{a+2}{27}+\frac{b+2}{27}+\frac{1}{9}>=4\sqrt[4]{\frac{\left(a+2\right)\left(b+2\right)}{27.27.9}.\frac{a^4}{\left(a+2\right)\left(b+2\right)}}...\)

\(>=\frac{4}{9}a\)

Tương tự

\(=>VT>=\frac{4}{9}\left(a+b+c\right)-\frac{3}{9}-2\left(\frac{a+2}{9}+\frac{b+2}{9}+\frac{c+2}{9}\right)=\frac{1}{3}.\)

Dấu "="xảy ra khi a=b=c=1

Ta có: \(a+b+c\ge3\sqrt[3]{abc}\)

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

\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)(1)

Ta có: \(\left(a-b\right)^3+\left(b-c\right)^2+\left(c-a\right)^3\)

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

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

\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Ta có: \(a-b+b-c+c-a\ge3\sqrt[3]{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(\Leftrightarrow0\ge\sqrt[3]{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(\Leftrightarrow0\ge3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(\Leftrightarrow9abc\ge9abc+3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)(2)

Từ (1), (2) ta có: \(\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc+3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc+\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)

Dấu "=" xảy ra khi \(a=b=c\)