ko khó nhưng mà bn đăng từng câu 1 hộ mk mk giải giúp cho

gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)

=> Thay vào thì     \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)

\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)

Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào

=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)

=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)

=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\) 

Đặt: \(\sqrt{a}=x;\sqrt{b}=y;\sqrt{c}=z\)

=>     \(P=\frac{xy}{z^2+3xy}+\frac{yz}{x^2+3yz}+\frac{zx}{y^2+3zx}\)

=>     \(3P=\frac{3xy}{z^2+3xy}+\frac{3yz}{x^2+3yz}+\frac{3zx}{y^2+3zx}=1-\frac{z^2}{z^2+3xy}+1-\frac{x^2}{x^2+3yz}+1-\frac{y^2}{y^2+3zx}\)

Ta sẽ CM: \(3P\le\frac{9}{4}\)<=> Cần CM: \(\frac{x^2}{x^2+3yz}+\frac{y^2}{y^2+3zx}+\frac{z^2}{z^2+3xy}\ge\frac{3}{4}\)

Có:    \(VT\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)

Ta sẽ CM: \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{3}{4}\)

<=> \(4\left(x+y+z\right)^2\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(4\left(x^2+y^2+z^2\right)+8\left(xy+yz+zx\right)\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(x^2+y^2+z^2\ge xy+yz+zx\)

Mà đây lại là 1 BĐT luôn đúng => \(3P\le\frac{9}{4}\)=> \(P\le\frac{3}{4}\)

Vậy P max \(=\frac{3}{4}\)<=> \(a=b=c\)

Ta có:

\(ab+bc+ca\le\dfrac{1}{3}\left(a+b+c\right)^2=3\)

\(\Rightarrow\dfrac{a}{\sqrt{a^2+3}}\le\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

Tương tự:

\(\dfrac{b}{\sqrt{b^2+3}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+3}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)

Cộng vế:

\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{c}{a+c}+\dfrac{a}{a+c}\right)=\dfrac{3}{2}\)

\(P_{max}=\dfrac{3}{2}\) khi \(a=b=c=1\)

Áp dụng BĐT cosi, ta có

\(\sqrt{3a+1}=\dfrac{1}{2}\sqrt{4\left(3a+1\right)}\le\dfrac{1}{2}.\dfrac{4+3a+1}{2}=\dfrac{3a+5}{4}\)

CMTT, ta có \(\sqrt{3b+1}\le\dfrac{3b+5}{4};\sqrt{3c+1}\le\dfrac{3c+5}{4}\)

Từ đó suy ra \(K\le\dfrac{3\left(a+b+c\right)+15}{4}=6\)

Dấu "=" xảy ra khi a=b=c=1

Vậy...

ta có BĐT \(\sqrt{3a+1}\ge\dfrac{a\left(\sqrt{10}-1\right)}{3}+1\)

\(\Leftrightarrow a\left(3-a\right)\ge0đúng\forall a\)

CMRTT, ta có

\(\sqrt{3b+1}\ge\dfrac{b\left(\sqrt{10}-1\right)}{3}+1\)

\(\sqrt{3c+1}\ge\dfrac{c\left(\sqrt{10}-1\right)}{3}+1\)

Do đó \(K\ge\dfrac{\left(a+b+c\right)\left(\sqrt{10}-1\right)}{3}+3=\sqrt{10}+2\)

Dấu "=" xảy ra khi a=3, b=c=0

Vậy...

Do vai trò của 3 biến là như nhau, ko mất tính tổng quát, giả sử \(a\ge b\ge c\)

\(\Rightarrow\) Theo BĐT Chebyshev:

\(3\left(a^3+b^3+c^3\right)\ge\left(a^2+b^2+c^2\right)\left(a+b+c\right)\) (1)

Bunhiacopxki:

\(\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2\le2\left(a^2+b^2+c^2\right)\left(a+b+c\right)\le6\left(a^3+b^3+c^3\right)\)

Nên ta chỉ cần chứng minh:

\(\left(a^3+b^3+c^3\right)^2\ge6\left(a^3+b^3+c^3\right)\)

\(\Leftrightarrow a^3+b^3+c^3\ge6\)

Hiển nhiên đúng do: \(a^3+b^3+c^3\ge3abc=6\)

1/ \(Q=\frac{\left(2-\sqrt{a}\right)\left(\sqrt{a}+3\right)}{\sqrt{a}+3}=2-\sqrt{a}\)

Do \(\sqrt{a}\ge0\Rightarrow2-\sqrt{a}\le2\Rightarrow Q_{max}=2\) khi \(a=0\)

2/

\(N=\sqrt{a+b+2\sqrt{\left(a+b\right)c}+c}+\sqrt{a+b-2\sqrt{\left(a+b\right)c}+c}\)

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

\(=\sqrt{a+b}+\sqrt{c}+\left|\sqrt{a+b}-\sqrt{c}\right|\)

TH1: Nếu \(a+b\ge c\Rightarrow\sqrt{a+b}-\sqrt{c}\ge0\)

\(\Rightarrow Q=\sqrt{a+b}+\sqrt{c}+\sqrt{a+b}-\sqrt{c}=2\sqrt{a+b}\)

TH2: Nếu \(a+b< c\Rightarrow\sqrt{a+b}-\sqrt{c}< 0\)

\(\Rightarrow Q=\sqrt{a+b}+\sqrt{c}+\sqrt{c}-\sqrt{a+b}=2\sqrt{c}\)

Dễ thấy theo AM - GM ta có:

\(P\ge3\sqrt[3]{\sqrt{\frac{a+b}{c+ab}\cdot\sqrt{\frac{b+c}{a+bc}}\cdot\sqrt{\frac{c+a}{b+ca}}}}\)

Ta cần chứng minh \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(c+ab\right)\left(a+bc\right)\left(b+ca\right)\)

Mặt khác theo AM - GM:

\(\left(c+ab\right)\left(a+bc\right)\le\frac{\left(c+ab+a+bc\right)^2}{4}=\frac{\left(b+1\right)^2\left(a+c\right)^2}{4}\)

Tương tự thì:

\(\left(c+ab\right)\left(a+bc\right)\left(b+ca\right)\le\frac{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8}\)

Ta cần chứng minh:\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le8\)

Áp dụng tiếp AM - GM:

\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le\frac{\left(a+1+b+1+c+1\right)^3}{27}=8\)

Vậy ta có đpcm

Chuyên Phan năm nay :))

3.

\(5a^2+2ab+2b^2=\left(a^2-2ab+b^2\right)+\left(4a^2+4ab+b^2\right)\)

\(=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)

Tương tự \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)

\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)

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

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

\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)