\(P+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\) \(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)

Áp dụng Cauchy-Schwarz dạng phân thức:

\(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\ge\frac{9}{2\left(a+b+c\right)}\)

\(\Leftrightarrow P+3\ge\frac{9}{2}\Rightarrow P\ge\frac{3}{2}\)

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

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2b^2}{b^2c^2}}\ge\frac{2a}{c}\) ; \(\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge\frac{2c}{b}\) ; \(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{2b}{a}\)

Cộng vế với vế ta có đpcm

Dấu "=" xảy ra khi \(a=b=c\)

2. \(\frac{bc}{a}+\frac{ac}{b}\ge2\sqrt{\frac{bc.ac}{ab}}=2c\) ; \(\frac{ac}{b}+\frac{ab}{c}\ge2a\) ; \(\frac{bc}{a}+\frac{ab}{c}\ge2b\)

Cộng vế với vế ta có đpcm

Dấu "=" xảy ra khi \(a=b=c\)

Thay abc = 1 vào bđt cần chứng minh : 

\(a+b+c\ge\frac{a\left(bc+1\right)}{b\left(ac+1\right)}+\frac{b\left(ac+1\right)}{c\left(ab+1\right)}+\frac{c\left(ab+1\right)}{a\left(bc+1\right)}\)

\(\Leftrightarrow a\left(1-\frac{bc+1}{ac+1}\right)+b\left(1-\frac{ac+1}{ab+1}\right)+c\left(1-\frac{ab+1}{bc+1}\right)\ge0\)

\(\Leftrightarrow\frac{ac\left(a-b\right)}{ac+1}+\frac{ab\left(b-c\right)}{ab+1}+\frac{bc\left(c-a\right)}{bc+1}\ge0\)

\(\Leftrightarrow\frac{ac\left[-\left(c-a\right)-\left(b-c\right)\right]}{ac+1}+\frac{ab\left[-\left(a-b\right)-\left(c-a\right)\right]}{ab+1}+\frac{bc\left[-\left(b-c\right)-\left(a-b\right)\right]}{bc+1}\ge0\)

\(\Leftrightarrow\left[\frac{-ac\left(c-a\right)}{ac+1}-\frac{ab\left(c-a\right)}{ab+1}\right]+\left[-\frac{ac\left(b-c\right)}{ac+1}-\frac{bc\left(b-c\right)}{bc+1}\right]+\left[-\frac{ab\left(a-b\right)}{ab+1}-\frac{bc\left(a-b\right)}{bc+1}\right]\ge0\)

\(\Leftrightarrow-a\left(c-a\right)\left(c+b\right)\left(\frac{1}{ac+1}+\frac{1}{ab+1}\right)-c\left(b-c\right)\left(a+b\right)\left(\frac{1}{ac+1}+\frac{1}{bc+1}\right)-b\left(a-b\right)\left(a+c\right)\left(\frac{1}{ab+1}+\frac{1}{bc+1}\right)\ge0\)(1)

Đặt \(x=\frac{1}{ab+1},y=\frac{1}{bc+1},z=\frac{1}{ac+1}\)

Tiếp tục phân tích : \(-c\left(b-c\right)\left(a+b\right).x-b\left(a-b\right)\left(a+c\right).y=-c\left(a+b\right).x\left[-\left(c-a\right)-\left(a-b\right)\right]-b\left(a+c\right).y\left[-\left(b-c\right)-\left(c-a\right)\right]\)

\(=\left(c-a\right).\left[c\left(a+b\right)x+b\left(a+c\right)y\right]+c\left(a+b\right)\left(a-b\right).x+b\left(a+c\right)\left(b-c\right).y\)

Tới đây giả sử \(a\ge b\ge c>0\) là ra nhé :)

Mình nghĩ CM bằng BĐT Bunhiacopxky đã là chi tiết rồi nhưng nếu bạn muốn chi tiết hơn nữa thì thế này:

Xét hiệu:\(\left(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\right)(x+y+z)-(a+b+c)^2\)

\(=a^2+a^2.\frac{y}{x}+a^2.\frac{z}{x}+b^2+b^2.\frac{x}{y}+b^2.\frac{z}{y}+c^2+c^2.\frac{x}{z}+c^2.\frac{y}{z}-(a^2+b^2+c^2-2ab-2bc-2ac)\)

\(=(a^2.\frac{y}{x}+b^2.\frac{x}{y}-2ab)+(a^2.\frac{z}{x}+c^2.\frac{x}{z}-2ac)+(b^2.\frac{z}{y}+c^2.\frac{y}{z}-2bc)\)

\(=(a\sqrt{\frac{y}{x}}-b\sqrt{\frac{x}{y}})^2+(a\sqrt{\frac{z}{x}}-c\sqrt{\frac{x}{z}})^2+(b\sqrt{\frac{z}{y}}-c\sqrt{\frac{y}{z}})^2\geq 0\) với mọi $a,b,c,x,y,z>0$

Do đó:\(\left(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\right)(x+y+z)\geq (a+b+c)^2\)

\(\Rightarrow \frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq \frac{(a+b+c)^2}{x+y+z}\) (đpcm)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\right)(x+y+z)\geq (a+b+c)^2\)

\(\Rightarrow \frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq \frac{(a+b+c)^2}{x+z+y}\) (đpcm)

Dấu "=" xảy ra khi \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

\(VT=\frac{a}{c}+\frac{b}{a}+\frac{c}{b}=\frac{a^2}{ca}+\frac{b^2}{ab}+\frac{c^2}{bc}\ge\frac{\left(a+b+c\right)^2}{ab+bc+ca}\)

Ta cần chứng minh :

\(\frac{\left(a+b+c\right)^2}{ab+bc+ca}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\frac{3}{ab+bc+ca}+2\ge\frac{9}{a+b+c}\)

Đặt \(a+b+c=t\)

\(\Leftrightarrow\frac{6}{t^2-3}+2\ge\frac{9}{t}\)

\(\Leftrightarrow2t^3-6t+6t-9t^2+27\ge0\)

\(\Leftrightarrow\left(2t+3\right)\left(t-3\right)^2\ge0\) ( Đúng )

Vậy ...

Giả sử \(a\ge b\ge c\)

\(\Rightarrow3a^2\ge a^2+b^2+c^2=3\Rightarrow a\ge1\)

Do đó ta có : \(a+b+c\ge1\)

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

Áp dụng BĐT Cô-si ta có :

\(\left\{{}\begin{matrix}\frac{a}{c}+\frac{b}{a}+\frac{b}{a}\ge3\sqrt[3]{\frac{b^2}{ac}}=\frac{3b}{\sqrt[3]{abc}}\\\frac{b}{a}+\frac{c}{b}+\frac{c}{b}\ge3\sqrt[3]{\frac{c^2}{ab}}=\frac{3c}{\sqrt[3]{abc}}\\\frac{c}{b}+\frac{a}{c}+\frac{a}{c}\ge3\sqrt[3]{\frac{a^2}{bc}}=\frac{3a}{\sqrt[3]{abc}}\end{matrix}\right.\)

\(\Rightarrow\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge\frac{a+b+c}{\sqrt[3]{abc}}\)

Ta cần chứng minh : \(\frac{a+b+c}{\sqrt[3]{abc}}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\left(\frac{a+b+c}{3}\right)^2\ge\sqrt[3]{abc}\)

Luôn đúng vì : \(\left(\frac{a+b+c}{3}\right)^2\ge\frac{a+b+c}{3}\ge\sqrt[3]{abc}\)

Dấu \("="\) xảy ra khi \(a=b=c=1\)

Áp dụng BĐT Cauchy-Schwarz ta có:

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

\(\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\). Mà theo BĐT AM-GM ta có:

\(\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}=\frac{\left(a+b+c+d\right)^2}{2\left[\left(a+b\right)\left(c+d\right)+\left(a+c\right)\left(b+d\right)+\left(a+d\right)\left(b+c\right)\right]}\ge\frac{2}{3}\)

Đẳng thức xảy ra khi a=b=c=d

là \(\frac{2}{3}\) nha

a) Áp dụng BĐT Cauchy-Schwarz dạng Engel: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

Tương tự:\(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c};\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)

Cộng theo vế 3 BĐT trên rồi chia cho 2 ta thu được đpcm

Đẳng thức xảy ra khi \(a=b=c\)

b)Đặt \(a+b=x;b+c=y;c+a=z\). Cần chứng minh:

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)

Cách làm tương tự câu a.

c) \(VT=\Sigma_{cyc}\frac{1}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\Sigma_{cyc}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\le\frac{1}{16}\Sigma\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)

Đẳng thức xảy ra khi \(a=b=c=\frac{3}{4}\)

d) Em làm biếng quá anh làm nốt đi:P

Câu d : \(p=\frac{a+b+c}{2}\Rightarrow2p=a+b+c\)

Ta có : \(\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{2p-a-b}=\frac{4}{c}\)

Tương tự : \(\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{a}\)

\(\frac{1}{p-c}+\frac{1}{p-a}\ge\frac{4}{b}\)

\(\Leftrightarrow2\left(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\right)\ge\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)

\(\Leftrightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\left(đpcm\right)\)

Dấu \("="\) xảy ra khi nó là tam giác đều