Ta có: \(3\left(x^2+y^2+z^2\right)-\left(x+y+z\right)^2=\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\)

\(\Rightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\) nên với \(x,y,z>0\) ta có:

\(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\) áp dụng ta có:

\(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ca+c+2}}\le\sqrt{3\left(\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\right)}\)

Với: \(x,y>0\) ta có: \(x+y\ge2\sqrt{xy}\Rightarrow\left(x+y\right)^2\ge4xy\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

Áp dụng ta được:

 \(\frac{1}{ab+a+2}=\frac{1}{ab+1+a+1}=\frac{1}{ab+abc+a+1}=\frac{1}{ab\left(c+1\right)+\left(a+1\right)}\)

\(\le\frac{1}{4}\left(\frac{1}{ab\left(c+1\right)}+\frac{1}{a+1}\right)=\frac{1}{4}\left(\frac{abc}{ab\left(c+1\right)}+\frac{1}{a+1}\right)=\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}\right)\)

Vậy ta có: \(\frac{1}{ab+a+2}\le\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}\right)\)

Tương tự như trên ta có: \(\frac{1}{bc+b+2}\le\frac{1}{4}\left(\frac{a}{a+1}+\frac{1}{b+1}\right)\) và \(\frac{1}{ca+c+2}\le\frac{1}{4}\left(\frac{b}{b+1}+\frac{1}{c+1}\right)\) nên:

\(\Rightarrow\sqrt{3\left(\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\right)}\)

\(\le\sqrt{3.\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}+\frac{a}{a+1}+\frac{1}{b+1}+\frac{b}{b+1}+\frac{1}{c+1}\right)}=\frac{3}{2}\)

Vậy \(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ca+c+2}}\le\frac{3}{2}\left(đpcm\right)\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)

Đặt \(\left(a;b;c\right)=\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)\). BĐT quy về:\(\Sigma_{cyc}\frac{\sqrt{yz}}{\sqrt{xy+xz+2yz}}\le\frac{3}{2}\)

Áp dụng liên hoàn BĐT Cô si:

\(VT=\Sigma_{cyc}\sqrt{\frac{yz}{\left(xy+yz\right)+\left(xz+yz\right)}}\le\Sigma_{cyc}\sqrt{\frac{yz}{4}\left(\frac{1}{xy+yz}+\frac{1}{xz+yz}\right)}\)

\(=\frac{1}{2}\Sigma_{cyc}\sqrt{1\left(\frac{yz}{xy+yz}+\frac{yz}{xz+yz}\right)}\le\frac{1}{4}\Sigma_{cyc}\left(1+\frac{yz}{xy+yz}+\frac{yz}{xz+yz}\right)=\frac{3}{2}\)

Áp dụng BĐT bunhiacopxki cho 2 bộ số \(\left(\sqrt{a}.\sqrt{b+c};\sqrt{b}.\sqrt{d+c};\sqrt{c}.\sqrt{d+a};\sqrt{d}.\sqrt{a+b}\right)\)

và \(\left(\frac{\sqrt{a}}{\sqrt{b+c}};\frac{\sqrt{b}}{\sqrt{d+c}};\frac{\sqrt{c}}{\sqrt{d+a}};\frac{\sqrt{d}}{\sqrt{a+b}}\right)\), ta được:

\(\left[a\left(b+c\right)+b\left(d+c\right)+c\left(d+a\right)+d\left(a+b\right)\right]\)\(\left(\frac{a}{b+c}+\frac{b}{d+c}+\frac{c}{a+d}+\frac{d}{a+b}\right)\)\(\ge\left(a+b+c+d\right)^2\)

\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{d+c}+\frac{c}{a+d}+\frac{d}{a+b}\)\(\ge\frac{\left(a+b+c+d\right)^2}{ab+ac+bd+bc+cd+ac+ad+bd}\)(1)

Ta có \(\left(a+b+c+d\right)^2\ge2\left(ab+ac+bc+bd+cd+ac+ad+bd\right)\)

\(\Leftrightarrow\left(a-c\right)^2+\left(b-d\right)^2\ge0\)(luôn đúng)

Do đó: \(\left(a+b+c+d\right)^2\ge2\left(ab+ac+bc+bd+cd+ac+ad+bd\right)\)(2)

Từ (1) và (2) suy ra ĐPCM

Dấu "=" xảy ra khi và chỉ khi a=b=c=d

Áp dụng BĐT : \(\frac{1}{xy}\ge\frac{4}{\left(x+y\right)^2}\)với x,y > 0

Ta có : \(\frac{a}{b+c}+\frac{c}{d+a}=\frac{a^2+ad+bc+c^2}{\left(b+c\right)\left(a+d\right)}\ge\frac{4\left(a^2+ad+bc+c^2\right)}{\left(a+b+c+d\right)^2}\)

Tương tự : \(\frac{b}{c+d}+\frac{d}{a+b}\ge\frac{4\left(b^2+ab+cd+d^2\right)}{\left(a+b+c+d\right)^2}\)

\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge\frac{4\left(a^2+b^2+c^2+d^2+ad+bc+ab+cd\right)}{\left(a+b+c+d\right)^2}\)

Cần chứng minh : \(\frac{a^2+b^2+c^2+d^2+ad+bc+ab+cd}{\left(a+b+c+d\right)^2}\ge\frac{1}{2}\)

\(\Leftrightarrow2\left(a^2+b^2+c^2+d^2+ad+bc+ab+cd\right)\ge\left(a+b+c+d\right)^2\)

\(\Leftrightarrow\left(a-c\right)^2+\left(b-d\right)^2\ge0\)

Dấu "=" xảy ra khi a = c ; b = d

Vậy ....

1/ BĐT \(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2\right)+4abc\ge104=\frac{13}{27}\left(a+b+c\right)^3\)

Hay: \(27\left(a+b+c\right)\left(a^2+b^2+c^2\right)+108abc\ge13\left(a+b+c\right)^3\)

\(VT-VP=2\left[6\left\{\Sigma_{cyc}a^3+3abc-\Sigma_{cyc}ab\left(a+b\right)\right\}+\left(a^3+b^3+c^3-3abc\right)\right]\ge0\)

(đúng theo BĐT Schur bậc 3 và Cô si cho 3 số dương)

Đẳng thức xảy ra khi a = b = c = 2

tth_new trả lời nốt luôn đi 

đkxđ : \(x,y,z\ge\frac{1}{4}\)

Ta có : 

\(x-z=\sqrt{4z-1}-\sqrt{4x-1}=\frac{4\left(z-x\right)}{\sqrt{4z-1}+\sqrt{4x-1}}=-\frac{4\left(x-z\right)}{\sqrt{4z-1}+\sqrt{4x-1}}\)

\(\Rightarrow\left(x-z\right)\left(1+\frac{4}{\sqrt{4z-1}+\sqrt{4x-1}}\right)=0\)

Dễ thấy \(1+\frac{4}{\sqrt{4z-1}+\sqrt{4x-1}}>0\)nên x - z = 0 hay x = z

Tương tự : x = y

Suy ra : x = y = z

Thay vào đầu bài, ta có : \(2x=\sqrt{4x-1}\Rightarrow4x^2=4x-1\Rightarrow x=\frac{1}{2}\)

Vậy x = y = z = \(\frac{1}{2}\)

Áp dụng BĐT Cô-si cho 3 số dương, ta có :

\(\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(a+c\right)}\ge3\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\)

Cần chứng minh : \(\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\ge\frac{9}{2\left(a+b+c\right)^2}\)

hay \(8\left(a+b+c\right)^6\ge729abc\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

Thật vậy, ta có : \(\left(a+b+c\right)^3\ge\left(3\sqrt[3]{abc}\right)^3=27abc\)

\(8\left(a+b+c\right)^3=\left(2\left(a+b+c\right)\right)^3=\left(a+b+b+c+a+c\right)^3\)

\(\ge\left(3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\right)^3=27\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

Nhân từng vế 2 bất đẳng thức trên, ta được đpcm

Dấu "=" xảy ra khi a = b = c 

Vậy ...

2. Áp dụng BĐT Cô-si cho 3 số không âm, ta có : 

\(B\ge3\sqrt[3]{\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(a^3+c^3+1\right)}}\)

Ta có : \(a^3+b^3+1\ge3\sqrt[3]{a^3b^3}=3ab\Rightarrow\sqrt{a^3+b^3+1}\ge\sqrt{3ab}\)

Tương tự : ....

\(\Rightarrow\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(c^3+a^3+1\right)}\ge\sqrt{27a^2b^2c^2}=\sqrt{27}\)

\(\Rightarrow B\ge3\sqrt[3]{\sqrt{27}}=3\sqrt{3}\)

Vậy GTNN của B là \(3\sqrt{3}\)khi a = b = c = 1

Áp dụng BĐT Cô-si dạng Engel,ta có :

\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}\)

Mà \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le x+y+z\)

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

Dấu "=" xảy ra khi x = y = z = \(\frac{3}{2}\)

nhầm sửa x = y = z = 1 nha

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