Câu c quen thuộc, chém trước:

Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)

Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)

Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)

\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)

Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)

\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)

Done.

Câu 1 chuyên phan bội châu

câu c hà nội

câu g khoa học tự nhiên

câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ

câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)

Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !

Nguyễn Ngọc Lộc , ?Amanda?, Phạm Lan Hương, Akai Haruma, @Trần Thanh Phương, @Nguyễn Việt Lâm,

@tth_new

Giúp em vs ạ! Thanks nhiều ạ

Cái c là \(\dfrac{2}{\sqrt{1+c^2}}\) ạ

\(P=\dfrac{2-\left(1+a^2\right)}{1+a^2}+\dfrac{2-\left(1+b^2\right)}{1+b^2}+\dfrac{2}{\sqrt{1+c^2}}\)

\(P=2\left(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{\sqrt{1+c^2}}\right)-2\) 

Từ điều kiện \(ab+bc+ca=1\), đặt \(\left\{{}\begin{matrix}a=tanx\\b=tany\\c=tanz\end{matrix}\right.\) với \(x+y+z=\dfrac{\pi}{2}\)

Xét \(Q=\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{\sqrt{1+c^2}}=\dfrac{1}{1+tan^2x}+\dfrac{1}{1+tan^2y}+\dfrac{1}{\sqrt{1+tan^2z}}\)

\(Q=cos^2x+cos^2y+cosz=1+\dfrac{1}{2}\left(cos2x+cos2y\right)+cosz\)

\(=1+cos\left(x+y\right)cos\left(x-y\right)+cosz\le1+cos\left(x+y\right)+cosz\)

\(=1+cos\left(\dfrac{\pi}{2}-z\right)+cosz=1+sinz+cosz=1+\sqrt{2}sin\left(z+\dfrac{\pi}{4}\right)\le1+\sqrt{2}\)

\(\Rightarrow P\le2\left(1+\sqrt{2}\right)-2=2\sqrt{2}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=y=\dfrac{\pi}{8}\\z=\dfrac{\pi}{4}\end{matrix}\right.\) \(\Rightarrow\left(a;b;c\right)=\left(\sqrt{2}-1;\sqrt{2}-1;1\right)\)

#Max: Giả sử z=max{x, y, z} \(\Rightarrow z\ge2\). Ta chứng minh BĐT sau:

\(x^2+y^2+z^2+xyz\le\dfrac{\left(x+y\right)^2}{2}+z^2+\dfrac{\left(x+y\right)^2z}{4}\)

\(\Leftrightarrow\dfrac{\left(x-y\right)^2}{4}\left(z-2\right)\ge0\) ( đúng ) (*)

Do đó \(VT\le\dfrac{\left(6-z\right)^2}{2}+z^2+\dfrac{z\left(6-z\right)^2}{4}=f\left(z\right)\) với \(z\in\left[2;3\right]\)

\(f'\left(z\right)=\left(6-z\right).\left(-1\right)+2z+\dfrac{1}{4}.\left[\left(6-z\right)^2+z.2\left(z-6\right)\right]\)

\(=\dfrac{3}{4}z^2-3z+3=\dfrac{3}{4}\left(z-2\right)^2\ge0\).Suy ra \(f\left(z\right)\le f\left(3\right)=\dfrac{81}{4}\)

Dấu = đạt được tại \(x=y=\dfrac{3}{2},z=3\) và các hoán vị

#Min: Để ý (*), ta giả sử z=Min{x, y, z} thì \(z\le2\). Do đó ta lại có

\(VT\ge f\left(z\right)\) với \(z\in\left[0;2\right]\). Vì f(z) vẫn đồng biến / R nên min sẽ đạt được tại z=0 và bằng 18

Dấu = đạt được tại x=y=3, z=0 và các hoán vị

\(P=\dfrac{a}{a^2+bc}+\dfrac{b}{b^2+ca}+\dfrac{c}{c^2+ab}\)

\(\le\dfrac{a}{2a\sqrt{bc}}+\dfrac{b}{2b\sqrt{ca}}+\dfrac{c}{2c\sqrt{ab}}\)

\(=\dfrac{a\sqrt{bc}}{2abc}+\dfrac{b\sqrt{ca}}{2abc}+\dfrac{c\sqrt{ab}}{2abc}\)

\(\le\dfrac{2a^2+b^2+c^2}{8abc}+\dfrac{2b^2+a^2+c^2}{8abc}+\dfrac{2c^2+b^2+a^2}{8abc}\)

\(=\dfrac{4\left(a^2+b^2+c^2\right)}{8abc}=\dfrac{1}{2}\)

\(P=\dfrac{\sqrt{a-2}}{a}+\dfrac{\sqrt[3]{b-3}}{b}+\dfrac{\sqrt[4]{c-6}}{c}\)

\(=\dfrac{\sqrt{\left(a-2\right).2}}{a\sqrt{2}}+\dfrac{\sqrt[3]{\left(b-3\right).\dfrac{3}{2}.\dfrac{3}{2}}}{b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{\sqrt[4]{\left(c-6\right).2.2.2}}{c\sqrt[3]{8}}\)

\(\le\dfrac{a-2+2}{2a\sqrt{2}}+\dfrac{b-3+\dfrac{3}{2}+\dfrac{3}{2}}{3b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{c-6+2+2+2}{4c\sqrt[4]{8}}\)

\(=\dfrac{a}{2a\sqrt{2}}+\dfrac{b}{3b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{c}{4c\sqrt[4]{8}}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{\dfrac{9}{4}}}+\dfrac{1}{4\sqrt[4]{8}}\)

Vậy \(P_{max}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{\dfrac{9}{4}}}+\dfrac{1}{4\sqrt[4]{8}}\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-2=2\\b-3=\dfrac{3}{2}\\c-6=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=\dfrac{9}{2}\\c=8\end{matrix}\right.\)

\(P=\dfrac{bc\sqrt{a-2}+ac\sqrt[3]{b-3}+ab\sqrt[4]{c-6}}{abc}\)

\(=\dfrac{\sqrt{a-2}}{a}+\dfrac{\sqrt[3]{b-3}}{b}+\dfrac{\sqrt[4]{c-6}}{c}\)

Áp dụng BĐT AM-GM ta có:

\(=\dfrac{\sqrt{2\left(a-2\right)}}{\sqrt{2}a}+\dfrac{\sqrt[3]{2\left(b-3\right)}}{\sqrt[3]{2}b}+\dfrac{\sqrt[4]{2\left(c-6\right)}}{\sqrt[4]{2}c}\)

\(\le\dfrac{\dfrac{2+a-2}{2}}{\sqrt{2}a}+\dfrac{\dfrac{2+b-3+1}{3}}{\sqrt[3]{2}b}+\dfrac{\dfrac{2+c-6+1+1+1+1}{4}}{\sqrt[4]{2}c}\)

\(=\dfrac{\dfrac{a}{2}}{\sqrt{2}a}+\dfrac{\dfrac{b}{3}}{\sqrt[3]{2}b}+\dfrac{\dfrac{c}{4}}{\sqrt[4]{2}c}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{2}}+\dfrac{1}{4\sqrt[4]{2}}\)

+) Bài bất đẳng thức:

\(\dfrac{2017a-a^2}{bc}=\dfrac{\left(a+b+c\right)a-a^2}{bc}=\dfrac{ab+ca}{bc}=\dfrac{a}{c}+\dfrac{a}{b}\left(1\right)\)

Tương tự: \(\left\{{}\begin{matrix}\dfrac{2017b-b^2}{ca}=\dfrac{b}{a}+\dfrac{b}{c}\left(2\right)\\\dfrac{2017c-c^2}{ab}=\dfrac{c}{a}+\dfrac{c}{b}\left(3\right)\end{matrix}\right.\)

\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow\dfrac{2017a-a^2}{bc}+\dfrac{2017b-b^2}{bc}+\dfrac{2017c-c^2}{ab}=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)

\(\sqrt{2}\left(\sum\sqrt{\dfrac{2017-a}{a}}\right)=\sqrt{2}\left(\sum\sqrt{\dfrac{\left(a+b+c\right)-a}{a}}\right)=\sqrt{2}\left(\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}+\sqrt{\dfrac{a+b}{2}}\right)\)

Bất đẳng thức cần chứng minh tương đương với:

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

*Có: \(\sqrt{2.\dfrac{a+b}{c}}+\sqrt{2.\dfrac{b+c}{a}}+\sqrt{2.\dfrac{c+a}{b}}\le\dfrac{2+\dfrac{a+b}{c}}{2}+\dfrac{2+\dfrac{b+c}{a}}{2}+\dfrac{2+\dfrac{c+a}{b}}{2}=3+\dfrac{\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}}{2}\)

Ta chỉ cần chứng minh:

\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge3+\dfrac{\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}}{2}\)

hay \(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\) (cái này chị tự chứng minh nhé)

Anh Trần Tuấn Hoàng giỏi BĐT quá nhỉ

Áp dụng BĐT B.C.S ta có

\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}\)

mặt khác do \(a+b+c\le3\Rightarrow\dfrac{9}{\left(a+b+c\right)^2}\ge1\)

\(\Rightarrow\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}\ge1\)(*)

ta lại có \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}\le3\)

\(\Rightarrow\dfrac{2007}{ab+bc+ac}\ge\dfrac{2007}{3}=669\)(**)

lấy (*)+(**) vế theo vế ta được

\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{2009}{ab+bc+ac}\ge669+1=670\left(dpcm\right)\)

Đặt \(\dfrac{b}{c}=x\)

Ta có: \(\left\{{}\begin{matrix}ab+bc=2c^2\\2a\le c\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{c}.x+x=2\\\dfrac{a}{c}\le\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{2-x}{x}\\\dfrac{2-x}{x}\le\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{2-x}{x}\\x\ge\dfrac{4}{3}\end{matrix}\right.\)

Ta lại có:

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

\(=\dfrac{\dfrac{2-x}{x}}{\dfrac{2-x}{x}-x}+\dfrac{x}{x-1}+\dfrac{1}{1-\dfrac{2-x}{x}}\)

\(=\dfrac{3x^2+8x-4}{2x^2+2x-4}\)

\(=\dfrac{27}{5}+\dfrac{39x^2+14x-88}{2x^2+2x-4}=\dfrac{27}{5}+\dfrac{\left(3x-4\right)\left(13x+22\right)}{2\left(x-1\right)\left(x+2\right)}\ge\dfrac{27}{5}\)

Vậy GTNN là \(\dfrac{27}{5}\) dấu = xảy ra khi \(x=\dfrac{4}{3}\)