Bài 1:a,b,c ba cạnh tam giác => a,b,c dương

\(\left\{{}\begin{matrix}a+c>b\\a+b>c\\b+c>a\end{matrix}\right.\) ta có: \(\dfrac{x}{y}< \dfrac{x+p}{y+p}\forall_{x,y,p>0\&x< y}\)

\(VT=\dfrac{a}{a+b}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+c}{a+b}+\dfrac{b}{c+a}< \dfrac{a+c+c}{a+b+c}+\dfrac{b+b}{a+b+c}=\)

\(=\dfrac{a+b+c+b+c}{a+b+c}< \dfrac{\left(a+b+c\right)+\left(A+b+c\right)}{a+b+c}< \dfrac{2\left(b+a+c\right)}{a+b+c}=2=VP\)

p/s: đề sao làm vậy:

mình nghi đề phải thế này: \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\) cách làm đơn giản hơn

Ta có \(a+b+c=abc\Leftrightarrow\dfrac{a+b+c}{abc}=1\) \(\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

Lại có \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

\(\Leftrightarrow2^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) (đpcm)

Bài 2:

Áp dụng BĐT: \(x^2+y^2+z^2\ge xy+yz+xz\), ta có:

\(a^4+b^4+c^4\ge a^2b^2+b^2c^2+a^2c^2\) (1)

Lại áp dụng tương tự ta có:

\(\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2\ge ab^2c+abc^2+a^2bc\)

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

Từ (1) và (2) suy ra:

\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

Bài 1:

Áp dụng BĐT Cô -si, ta có:

\(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\sqrt[3]{\dfrac{a^2}{b^3}.\dfrac{1}{a}.\dfrac{1}{a}}=\dfrac{3}{b}\)

\(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\sqrt[3]{\dfrac{b^2}{c^3}.\dfrac{1}{b}.\dfrac{1}{b}}=\dfrac{3}{c}\)

\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)

Cộng vế theo vế ta được:

\(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{a^2}{a^3}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

p/s: không chắc lắm, có gì sai xót xin giúp đỡ

Ta có đánh giá sau với a không âm:

\(\dfrac{a}{1+a^2}\le\dfrac{36a+3}{50}\)

Thật vậy, BĐT tương đương:

\(\left(36a+3\right)\left(a^2+1\right)\ge50a\)

\(\Leftrightarrow\left(3a-1\right)^2\left(4a+3\right)\ge0\) (luôn đúng)

Tương tự: \(\dfrac{b}{1+b^2}\le\dfrac{36b+3}{50}\) ; \(\dfrac{c}{1+c^2}\le\dfrac{36c+3}{50}\)

Cộng vế: \(VT\le\dfrac{36\left(a+b+c\right)+9}{50}=\dfrac{9}{10}\)

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

Ta chứng minh bđt phụ \(\dfrac{a}{1+a^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\)

Thật vậy bđt trên \(\Leftrightarrow\dfrac{-3a^2+10a-3}{10\left(1+a^2\right)}-\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\le0\)

\(\Leftrightarrow\left(a-\dfrac{1}{3}\right)\left[\dfrac{3\left(3-a\right)}{10\left(1+a^2\right)}-\dfrac{18}{25}\right]\le0\)

\(\Leftrightarrow-\dfrac{36\left(a-\dfrac{1}{3}\right)^2\left(\dfrac{3}{4}+a\right)}{50\left(1+a^2\right)}\le0\) ( luôn đúng với mọi \(a\)\(\ge\)0)

Tương tự cũng có:\(\dfrac{b}{1+b^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(b-\dfrac{1}{3}\right)\); \(\dfrac{c}{1+c^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(c-\dfrac{1}{3}\right)\)

Cộng vế với vế => VT\(\le\dfrac{9}{10}+\dfrac{18}{25}\left(a+b+c-1\right)=\dfrac{9}{10}\)

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

a;b;c dương k bn

Ta có:

\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\)

=>\(\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)=a+b+c\)

=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+c\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)+b\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)+a\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)=a+b+c\)

=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)

=>\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)(đpcm)

Ta có: \(pt\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{b\left(c+a\right)}{c+a}+\dfrac{c\left(a+b\right)}{a+b}=a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{ab}{b+c}+\dfrac{ac}{b+c}+\dfrac{bc}{c+a}+\dfrac{ba}{c+a}+\dfrac{ca}{a+b}+\dfrac{cb}{a+b}=a+b+c\)

\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c\)

\(\Leftrightarrow\) đpcm

Từ cái sau suy ra cái trước thì còn dễ hơn, đề này tui mà chưa làm kiểu kia sao bk làm :)

có \(\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{a+c}+\dfrac{c\left(a+b+c\right)}{a+b}\)=a=b+c

=>\(\dfrac{a^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+b\left(a+c\right)}{a+c}+\dfrac{c^2+c\left(a+b\right)}{a+b}\)=a+b+c

=>\(\dfrac{a^2}{b+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{b^2}{a+c}+\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c^2}{a+b}+\dfrac{c\left(a+b\right)}{a+b}\)=a+b+c

=>\(\dfrac{a^2}{b+c}+a+\dfrac{b^2}{a+c}+b+\dfrac{c^2}{a+b}+c=a+b+c\)

tự làm tiếp

Lời giải:

Ta có:

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)

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

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

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

\(\Leftrightarrow \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)

Ta có đpcm.

Bài 1:

Áp dụng t.c của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)