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 AM - GM, ta có:

\(\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\)

\(\ge\dfrac{3\sqrt{a^3b^3c^3}}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{a^2+c^2}{b^2+\dfrac{a^2+c^2}{2}}\)

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

\(=\dfrac{3}{2}+2\times\left[\dfrac{a^2+b^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\dfrac{b^2+c^2}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}+\dfrac{c^2+a^2}{\left(b^2+c^2\right)+\left(b^2+a^2\right)}\right]\) (1)

Đặt \(\left\{{}\begin{matrix}a^2+b^2=x\\b^2+c^2=y\\c^2+a^2=z\end{matrix}\right.\), ta có:

\(\left(1\right)\Leftrightarrow\dfrac{3}{2}+2\times\left(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right)\)

\(\ge\dfrac{3}{2}+2\times\dfrac{3}{2}\) (Bất_đẳng_thức_Nesbitt)

\(=\dfrac{9}{2}\left(\text{đ}pcm\right)\)

Dấu "=" xảy ra khi a = b = c

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 \(a=b=c=\dfrac{1}{\sqrt{3}}\)

Ta có : \(\dfrac{1}{a^2+b^2}+\dfrac{1}{b^2+c^2}+\dfrac{1}{c^2+a^2}\le\dfrac{a^3+b^3+c^3}{2abc}+3\)

\(\Leftrightarrow\dfrac{a^2+b^2+c^2}{a^2+b^2}+\dfrac{a^2+b^2+c^2}{b^2+c^2}+\dfrac{a^2+b^2+c^2}{c^2+a^2}\le\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{b^3}{2abc}+3\)( vì \(a^2+b^2+c^2=1\) )

\(\Leftrightarrow3+\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\le\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}+3\)

\(\Leftrightarrow\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\le\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)

Mà theo bất đẳng thức cô-si , ta có : \(b^2+c^2\ge2bc\)\(\Rightarrow\dfrac{a^2}{b^2+c^2}\le\dfrac{a^2}{2bc}\)

Tương tự ta cũng có : \(\dfrac{b^2}{c^2+a^2}\le\dfrac{b^2}{2ca},\dfrac{c^2}{a^2+b^2}\le\dfrac{c^2}{2ab}\)

Cộng các bất đẳng thức trên lại với nhau ta được :

\(\Leftrightarrow\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\le\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)

Do đó bất đẳng thức ban đầu được chứng minh .

a) Áp dụng bất đẳng thức Schur với \(r=1\)

\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)

\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)

\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c\)

b) Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)

Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)

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

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

Dấu " = " xảy ra khi \(a=b=c\)

c) Ta có \(abc=ab+bc+ca\)

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

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0

\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)

Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)

\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0

\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )

Từ ( 1 ) và ( 2 )

\(\Rightarrow VT\le\dfrac{3}{16}\)

\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )

Câu 2 b sos là ra :D

\(BĐT\Leftrightarrow\Sigma_{cyc}\left(\frac{a^3-ab^2}{b^2}\right)\ge0\Leftrightarrow\Sigma_{cyc}\left(\frac{a\left(a-b\right)\left(a+b\right)}{b^2}-2\left(a-b\right)\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\left(a-b\right)\left(\frac{a^2-b^2+ab-b^2}{b^2}\right)\ge0\)\(\Leftrightarrow\Sigma_{cyc}\frac{\left(a-b\right)^2\left(a+2b\right)}{b^2}\ge0\)

BĐT cuối cùng đúng nên ta có Q.E.D.

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

Câu 3 có cách khác ạ,nhưng mà sao em thấy nó sai sai,vì dấu "=" không xảy ra!Mong mọi người check giúp ạ!

gt<=> \(abc=ab+bc+ca\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) (chia hai vế cho abc>0)

Dễ chứng minh \(\frac{1}{36}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{k}\right)\ge\frac{1}{a+b+c+d+e+k}\)

Áp dụng vào,ta có: \(VT\le\frac{6}{36}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{6}< \frac{3}{16}?!?\)

\(P\ge\dfrac{3abc}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{c^2+a^2}{b^2+\dfrac{c^2+a^2}{2}}\)

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

Đặt \(\left(a^2+b^2;b^2+c^2;a^2+c^2\right)=\left(x;y;z\right)\)

\(\Rightarrow P\ge\dfrac{3}{2}+2\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)=\dfrac{3}{2}+2\left(\dfrac{x^2}{xy+xz}+\dfrac{y^2}{yz+xy}+\dfrac{z^2}{xz+yz}\right)\)

\(P\ge\dfrac{3}{2}+\dfrac{2\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{3}{2}+\dfrac{3\left(xy+yz+zx\right)}{xy+yz+zx}=3+\dfrac{3}{2}=\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)

tịt

Lời giải:

Vì $1=a^2+b^2+c^2$ nên:

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

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

\(\leq 3+\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}\) (theo BĐT AM-GM)

\(=3+\frac{a^3+b^3+c^3}{2abc}\) (đpcm)

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