Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\) khi đó thu được \(xyz=1\)

Ta có:

\(\dfrac{1}{a^2\left(b+c\right)}=\dfrac{x^2}{\dfrac{1}{y}+\dfrac{1}{z}}=\dfrac{x^2yz}{y+z}=\dfrac{x}{y+z}\)

BĐT cần chứng minh được viết lại thành:\(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\left(\dfrac{x}{y+z}+1\right)+\left(\dfrac{y}{z+x}+1\right)+\left(\dfrac{z}{x+y}+1\right)\ge\dfrac{9}{2}\)

\(\Leftrightarrow\left(x+y+z\right)\left(\dfrac{1}{y+z}+\dfrac{1}{z+x}+\dfrac{1}{x+y}\right)\ge\dfrac{9}{2}\)

Đánh giá cuối cùng đúng theo BĐT Cauchy

Vậy BĐT được chứng minh. Đẳng thức xảy ra khi và chỉ khi  a = b = c = 1.

3) Biến đổi tương đương:

\(8\left(a^3+b^3+c^3\right)\ge\left(a+b\right)^3+\left(b+c\right)^3+\left(a+c\right)^3\) (1)

\(\Leftrightarrow\left(a^3+b^3\right)+\left(b^3+c^3\right)+\left(a^3+c^3\right)+6\left(a^3+c^3+b^3\right)\)

\(\ge\left(a^3+b^3\right)+\left(b^3+c^3\right)+\left(a^3+c^3\right)+3ab\left(a+b\right)+3bc\left(b+c\right)+3ac\left(a+c\right)\)

\(\Leftrightarrow2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\)

\(\Leftrightarrow\left[a^3+b^3-ab\left(a+b\right)\right]+\left[a^3+c^3-ac\left(a+c\right)\right]+\left[b^3+c^3-bc\left(b+c\right)\right]\ge0\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2+\left(a+c\right)\left(a-c\right)^2+\left(b+c\right)\left(b-c\right)^2\ge0\) luôn đúng do a, b, c > 0

=> (1) đúng

Dấu "=" xảy ra khi a = b = c

4) Ta có: a+b>c ; b+c>a; a+c>b

Xét \(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)

Tương tự: \(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c}\)

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

Vậy suy ra được điều phải chứng minh

2) Xét: \(\dfrac{a^2}{b^2+c^2}-\dfrac{a}{b+c}=\dfrac{a\left(ab+ac-b^2-c^2\right)}{\left(b^2+c^2\right)\left(b+c\right)}=\dfrac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b^2+c^2\right)\left(b+c\right)}\left(1\right)\)

Tương tự:

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

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

Cộng (1),(2),(3) vế theo vế ta được:

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

\(=ab\left(a-b\right)\left[\dfrac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\dfrac{1}{\left(a^2+c^2\right)\left(a+c\right)}\right]+ac\left(a-c\right)\left[\dfrac{1}{b^2+c^2\left(b+c\right)}-\dfrac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]+bc\left(b-c\right)\left[\dfrac{1}{\left(a^2+c^2\right)\left(a+c\right)}-\dfrac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]\)

Giả sử \(a\ge b\ge c>0\) thì các biểu thức trong ngoặc tròn, vuông không âm

=> đpcm

áp dụng bdt côsi \(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{3}{b}\)

tuông tu \(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{3}{c}\)

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

suy ra vt +\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

suy ra dpcm

dau = xay ra khi a=b=c

Từ bài toán này (mà bạn đã hỏi cách đây vài bữa):

cho a,b,c>0. Chứng minh rằng: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\) - Hoc24

Ta có: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\)

Do đó: \(VT\ge\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}\)

Lại có: \(\dfrac{a+b+c}{\sqrt[3]{abc}}\ge\dfrac{3\sqrt[3]{abc}}{\sqrt[3]{abc}}=3\)

Đặt \(\dfrac{a+b+c}{\sqrt[3]{abc}}=x\ge3\Rightarrow VT\ge x+\dfrac{1}{x}=\dfrac{x}{9}+\dfrac{1}{x}+\dfrac{8x}{9}\ge2\sqrt{\dfrac{x}{9x}}+\dfrac{8}{9}.3=\dfrac{10}{3}\) (đpcm)

bạn ơi, bài này sai đề rồi

Ta có: BĐT\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{1}{2}+\dfrac{b}{b+c}-\dfrac{1}{2}+\dfrac{c}{c+a}-\dfrac{1}{2}\ge0\)

\(\Leftrightarrow\dfrac{2a-\left(a+b\right)}{2\left(a+b\right)}+\dfrac{2b-\left(b+c\right)}{2\left(b+c\right)}+\dfrac{2c-\left(c+a\right)}{2\left(c+a\right)}\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-a+a-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+\dfrac{a-c}{2}\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{\left(b+c\right)\left(c+a\right)}\right)\ge0\)

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

Vậy BĐT luôn đúng với \(a\ge b\ge c>0\)