3/ Áp dụng bất đẳng thức AM-GM, ta có :

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)

\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)

Cộng 3 vế của BĐT trên ta có :

\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)

\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)

Do đó:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Bài 2:

Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:

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

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

\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

Áp dụng BĐT AG-GM:

\(\dfrac{a^3}{a^2+ab+b^2}\ge\dfrac{a^3}{a^2+\dfrac{a^2+b^2}{2}+b^2}=\dfrac{a^3}{\dfrac{3}{2}\left(a^2+b^2\right)}\)

Cmtt \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^3}{b^2+bc+c^2}\ge\dfrac{b^3}{\dfrac{3}{2}\left(b^2+c^2\right)}\\\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{c^3}{\dfrac{3}{2}\left(c^2+a^2\right)}\end{matrix}\right.\)

Cộng vế theo vế của bất đẳng thức:

\(\Leftrightarrow VT\ge\dfrac{2}{3}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\)

Tiếp tục áp dụng BĐT AG-GM:

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

Cmtt\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^3}{b^2+c^2}\ge b-\dfrac{c}{2}\\\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\end{matrix}\right.\)

Cộng vế theo vế

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

\(\dfrac{a^3}{a^2+ab+b^2}=a-\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^2.ab.b^2}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)

Tương tự và cộng lại ta sẽ có đpcm

Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)

\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)

CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)

Cộng vế theo vế:

\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

Áp dụng bđt AM-GM:

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

\("="\Leftrightarrow a=b=c\)

Hân đz đã đến :v giờ lm nha

Ta có: \(a^3=a\cdot a^2\)

\(\Rightarrow a^3+a\cdot b^2=a\cdot a^2+a\cdot b^2=a\left(a^2+b^2\right)\)

\(\Rightarrow\dfrac{a^3}{a^2+b^2}=\dfrac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\)(*)

Ta có: \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)

\(\Rightarrow\dfrac{ab^2}{a^2+b^2}\le\dfrac{ab^2}{2ab}=\dfrac{b}{2}\)

\(\Rightarrow\dfrac{a^3}{a^2+b^2}\ge a-\dfrac{b}{2}\)

Tương tự: \(\dfrac{b^3}{b^2+c^2}\ge b-\dfrac{c}{2}\); \(\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\)

Cộng 3 bđt trên ta có:

\(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\ge a+b+c-\dfrac{b}{2}-\dfrac{c}{2}-\dfrac{a}{2}=\dfrac{a+b+c}{2}\)

''='' xảy ra khi \(a=b=c\)

lát hông ai làm thì t lm cho :))

Áp dụng bđt cosi schwart ta có:

`VT>=(a+b+c)^2/(a+b+c+sqrt{ab}+sqrt{bc}+sqrt{ca})`

Dễ thấy `sqrt{ab}+sqrt{bc}+sqrt{ca}<a+b+c`

`=>VT>=(a+b+c)^2/(2(a+b+c))=(a+b+c)/2=3`

Dấu "=" `<=>a=b=c=1.`

Đặt vế trái BĐT là P

Ta có:

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

\(\Rightarrow P.\left(2ab+2bc+2ca\right)\ge1\)

\(\Rightarrow P\ge\dfrac{1}{2\left(ab+bc+ca\right)}\ge\dfrac{1}{2\left(a^2+b^2+c^2\right)}=\dfrac{1}{2}\) (đpcm)

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

\(\dfrac{a^3}{1+b}+\dfrac{1+b}{4}+\dfrac{1}{2}\ge3\sqrt[3]{\dfrac{a^3\left(1+b\right)}{8\left(a+b\right)}}=\dfrac{3a}{2}\)

\(\dfrac{b^3}{1+c}+\dfrac{1+c}{4}+\dfrac{1}{2}\ge\dfrac{3b}{2}\) ; \(\dfrac{c^3}{1+a}+\dfrac{1+a}{4}+\dfrac{1}{2}\ge\dfrac{3c}{2}\)

\(\Rightarrow VT+\dfrac{a+b+c}{4}+\dfrac{9}{4}\ge\dfrac{3}{2}\left(a+b+c\right)\)

\(\Rightarrow VT\ge\dfrac{5}{4}\left(a+b+c\right)-\dfrac{9}{4}\ge\dfrac{5}{4}.3\sqrt[3]{abc}-\dfrac{9}{4}=\dfrac{3}{2}\)